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ELEMENTS 


OF 


Conic  Sections 


AND 


Analytical  Geometry. 


BT 

JAMES  H.  COFFIN,  LL.D., 

LATB  FBOFESSOB  OF  MATHEMATICS  AND  PHT8ICS  IN  LAFAYETTE  COLLEGE,  ASD 
AUTHOR  OF  TREATISES  ON  SOLAR  AND  LUNAR  ECLIPSES,  ASTRO- 
NOMICAL TABLES,  THE  WINDS  OP  THE  GLOBE,  ETC. 


SIXTH     EDITION,      REVISED     AND     IMPROVED, 

BY 

SELDEN  J.  COFFIN,  Ph.  D. 

HOLLENBACK  PROFESSOR  OF  MATHEMATICS  IN  LAFAYETTE  COLLEGE. 


NEW    YORK: 

COLLINS    &    BROTHER,     PUBLISHERS, 

No.    414    BROADWAY. 


COFFIN'S  ECLIPSES.  Solar  and  Lunar  Eclipses  familiarly  illustrated 
and  explained,  with  the  method  of  calculating  them,  as  taught  in  the  New 
England  Colleges.  By  James  H.  Coffin,  LL.D.  ,  Professor  of  Mathematics 
and  Physics,  Lafayette  College,  Pa.    8vo.     Price,  $1.65. 

LIFE  OP  JAMES  H.  COFFIN",  LL.D.,  for  twenty -seven  years  Pro- 
fessor of  Mathematics  and  Astronomy  in  Lafayette  College,  Member  of  the 
National  Academy  of  Sciences,  and  author  of  "  The  Winds  of  the  Globe," 
etc.  By  Rev.  John  C.  Clyde,  Author  of  the  "History  of  the  Irish  Settle- 
ment," and  "Life  of  Rosbrugh,"  etc.    373  pp.,  12mo,  cloth.     $1.50. 

RECORD  OF  THE  MEN  OF  LAFAYETTE;  Brief  Biographical 
Sketches  of  all  the  Alumni  of  Lafayette  College.  By  Rev.  Selden  J. 
Coffin,  Ph.  D.,  with  Prof.  Owen's  Historical  Sketches  of  the  College. 
428  pp.,  8vo.,  illustrated.     $3.00. 

Copies  mailed,  postpaid,  on  receipt  of  price. 

COLLINS  &  BROTHER, 

414  Bkoadwat,  N.  T. 

GIFT 


COPTBIOHT,  1881,  BY  S.  J.  COPKN. 


CS-7 


PREFACE. 


The  following  treatise  has  been  prepared  to  meet  the  wants  of 
the  author,  in  the  instruction  of  his  classes.  He  has  felt  the  need 
of  a  work  on  the  Conic  Sections,  that  was  not,  on  the  one  hand,  so 
prolix  and  tedious  in  the  method  of  demonstration  as  to  render  the 
study  repulsive  to  the  student ;  nor,  on  the  other,  so  meager  as 
to  the  number  of  properties  discussed,  as  to  give  him  but  a  very 
imperfect  idea  of  the  interesting  features  of  these  curves,  and  ma- 
terially to  cripple  his  future  course  of  study,  which,  if  properly 
conducted,  requires  a  thorough  knowledge  of  them.  In  the  prepa- 
ation  of  this  work,  it  has  been  the  aim  to  avoid  both  these  defects : 
so  as,  on  the  one  hand,  to  render  it  as  full  and  complete  as  the  most 
thorough  works  in  use  upon  the  subject ;  and,  on  the  other,  to 
lighten  the  labor  >f  the  student,  by  simplifying  the  demonstrations 
without  rendering  them  less  rigid, — thus  giving  him  a  more  clear 
and  perfect  knowledge  of  the  properties  discussed,  and  at  the  same 
time  diminishing  the  size  of  the  book. 

The  properties  of  the  Conic  Sections  may  be  investigated  by 
either  of  two  quite  dissimilar  methods ;  each  of  which  has  its  pe- 
cjuliar  advantage?.  We  may  study  them  directly  from  the  figure 
itself,  in  the  same  manner  as  in  elementary  geometry;  and  this 
method,  which  Ms  called  the  geometrical,  has  the  advantage  of  af- 


M6414L87 


4  PREFACE. 

fording  a  more  clear  conception  of  the  properties  under  considera* 
tion.  Or  we  may,  after  the  method  invented  by  Descartes,  first 
represent  the  several  parts  of  the  figure  by  an  equation,  and  then 
proceed  in  our  investigations  by  pure  algebra.  This  method,  which 
is  called  the  analytical,'^  has  the  advantage  of  enabling  us  to  extend 
our  researches  far  beyond  what  we  could  otherwise  do  ;  just  as,  by 
the  aid  of  ordinary  algebra,  we  can  solve  questions  which  it  would 
be  impossible  to  solve  by  common  arithmetic.  The  former  method 
is  better  adapted  to  make  clear  and  sound  reasoners ;  the  latter, 
when  used  in  its  proper  sphere,  expert  and  finished  mathematicians. 
It  cannot,  however,  profitably  supersede  the  geometrical  method  in 
cases  to  which  the  latter  is  applicable.  Indeed,  as  just  hinted,  the 
analytical  method  is  to  geometry  what  algebra  is  to  common  arith- 
metic— valuable  as  an  aid,  but  absurd  as  a  substitute.  It  has  been 
sometimes  supposed  that  the  properties  of  the  Conic  Sections  could 
be  more  easily  investigated  by  the  analytical  method,  and  the  ex- 
ceedingly tedious  geometrical  demonstrations  that  we  find  in  some 
works,  certainly  afford  grounds  for  the  opinion.  But  it  need  not 
be  so.  All  the  leading  properties  can  be  demonstrated  with  equal 
ease,  and  greater  clearness,  by  the  geometrical  method,  while  it  is 
the  province  of  analytical  geometry  to  apply  them.  A  knowledge 
of  both  is,  therefore,  essential  to  a  perfect  course  in  mathematics. 

In  accordance  with  the  foregoing  views,  this  treatise  consists  of 
two  parts.  In  the  First  Part  the  various  properties  of  the  Conic 
Sections  are  demonstrated,  for  the  most  part  geometrically ;  and,  m 
the  Second,  the  student  is  taught  how  to  represent  lines,  curves 
and  surfaces  analytically y  and  to  solve  problems  relating  to  them. 


Often  called  the  French  method. 


PREFACE.  ft 

Our  definition  of  a  Conic  Section,  is  merely  an  extension  of  the 
common  definition  of  a  parabola,  and  is  recommended  by  the  follow- 
ing considerations : —  ^ 

I.  It  is  general,  belonging  to  each  of  the  three  curves.     The  more 

common  method  is,  to  define  the  ellipse,  parabola,  and  hyperbola 
as  three  distinct  curves.  They  are  called  conic  sections,  but  is  not 
the  student  left  in  the  dark  as  to  what  a  conic  section  is,  or  why 
these  curves  are  called  by  the  same  general  name  ? 

II.  By  thus  uniting  the  sections,  and  showing  that  instead  of 
being  three  different  curves,  they  are  merely  modifications  of  one 
and  the  same  curve — the  conic  section,  the  mind  of  the  student  is 
better  prepared  to  appreciate  the  analogies  that  he  finds  between 
them. 

III.  It  simplifies  the  demonstrations,  as  it  enables  us  at  the  outset 
to  prove  both  of  the  fundamental  propositions  of  other  treatises 
independently  of  each  other;  so  that  we  can  avail  ourselves  of 
either  at  pleasure  in  the  subsequent  demonstrations.  It  is  to  this 
fact  chiefly  that  many  of  the  demonstrations  owe  their  simplicity. 

With  a  view  to  keep  the  analogies  between  the  three  curves 
prominent  before  the  mind  of  the  student,  it  is  the  author's  practice 
with  his  classes  to  take  up  the  corresponding  propositions  in  con- 
nection, instead  of  following  the  order  of  the  book ;  and  for  the 
sake  of  convenience  in  giving  out  the  lessons,  they  are  numbered 
lilrke  in  the  second,  third,  and  fourth  chapters. 

The  subject  of  the  curvature  of  the  Conic  Sections  can  be  ais- 
cussed  to  better  advantage  by  the  aid  of  the  Differential  Calculus ; 
but  foi  the  benefit  of  those  who  are  not  acquainted  with  that  branch 


6  PREFACE. 

of  mathematics,  Chapter  V.  is  given  as  a  substitute.  The  last 
proposition  in  this  chapter,  and  the  last  two  in  Chapter  VI.,  discuss 
properties  not  treated  of  in  other  works  on  the  Conic  Sections,  but 
thought  to  be  important  from  their  applications  in  physical  as- 
tronomy. 

J.  H.  C. 


NOTE  TO  THE  REVISED  EDITION. 

I  HAVE  not  thought  it  possible  to  make  any  material  improve- 
ment upon  the  elegance  and  conciseness  of  the  author's  proofs  of 
Propositions  in  the  accompanying  treatise  on  Analytical  Geometry. 
I  have  added  a  number  of  original  Numerical  Exercises,  adapted  to 
illustrate  the  meaning  of  the  Propositions,  and  to  show  the  utility 
of  the  truths  taught. 

The  student  of  Conic  Sections  will  welcome  the  explanation  of 
geometrical  points,  made  by  the  introduction  of  numerous  refer- 
ences to  the  treatises  of  Professors  Loomis  and  Wentworth,  which 
are  given  in  addition  to  those  to  Legendre  and  Euclid  in  the  earlier 

editions. 

S.  J.  C. 

Lafayette  College,  September  1,  1881. 


Tfe 


CONTENTS 


CONIC    SECTIONS. 

CHAPTER  I. 
DEFnanoHs  and  General  Propositions 9 

CHAPTER  II. 
Or  THE  Ellipse , , 14 

CHAPTER  III. 
Of  the  Parabola 34 

CHAPTER  IV. 
Of  the  Hyperbola 44 

CHAPTER  V. 
Curvature  of  the  Conic  Sections 64 

CHAPTER  VI. 
Properties  peculiar  to  the  different  Conic  Sections 71 


ANALYTICAL    GEOMETRY. 

CHAPTER  I. 
Of  the  Point  and  Straight  Line  in  a  Plane 83 

CHAPTER  II. 
Of  Curves — Circle — Ellipse— Parabola — Hyperbola 94 


8  CONTENTS. 

rAOB 

CHAPTER  III. 
Of  Lines  in  Space 126 

CHAPTER  IV. 
Of  Plane  Surfaces 137 

CHAPTER  V. 
Of  Cubved  Surfaces 145 


^PFENDLX 153 

Practical  Exercises  in  Conic  Sections 159 

Numerical  Exercises  in  Analytical  Geometry 1 60 


PART    I. 


CONIC    SECTIONS. 


CHAPTER  I 


DEFINITIONS   AND    GENERAL   PROPOSITIONS. 


(1)  -4.  conic  section  is  a  curve,  the  distance  of  any  point  in  which 
from  a  given  point,  is  to  its  distance  from  a  given  straight  line  in  a 
given  ratio 

If  the  distance  to  the  point  he  less  than  to  the  line,  the  curve  is 
called  an  Ellipse ;  if  equal,  a  Parabola ;  and  if  greater,  an  HypeV' 
hola.^ 

For  example;  in  the  following  figure,  in  which  AB  represents 
the  given  line,  F  the  given  point,  and  R 
any  point  in  the  curve  MN,  if  the  ratio  of 
RS  to  RF  continues  the  same  wherever 
in  the  curve  the  point  R  is  taken,  the 
curve  is  a  conic  section,  and  is  an  ellipse, 
parabola,  or  hyperbola,  according  as  RF 
is  less,  equal  to,  or  greater  than  RS. 


'  If  the  distance  to  the  line  be  infinite,  the  curve  becomes  a  circle ;  and  if  the 
distance  to  the  point  be  infinite,  the  curve  becomes  a  straight  line. 

2 


10 


CONIC  SECTIONS. 


(2)  Prop.  I.     Problem. 

To  describe  a  conic  section. 

(Fig.  1.  Ellipse.) 
For  Parabola  and  Hyperbola  6t>e  page  153.       L 


Let  F  be  the  given  point,  AB  the  given  straight  line,  and  m :  n 
the  given  ratio. 

Through  F  draw  DE  parallel,  and  CX  perpendicular  to  AB,. 
each  of  indefinite  length.  From  FD  and  FE  cut  off  FM  and  FN, 
so  that  each  shall  be  to  FC  in  the  given  ratio ;  that  is,  FM  or 
FN  :  FC  ::  m  :  n.  Through  the  points  CM  and  C.N  draw  KL 
and  PQ  of  indefinite  length,  and  from  them  draw  a  series  of  per- 
pendiculars to  CX,  as  a.l,  h.2,  c.3,  4.1,  5.2,  6.3,  &c.  Take  the  length 
of  any  perpendicular  in  the  compasses,  and  with  one  foot  on  F, 
note  where  the  other  falls  on  that  perpendicular.  The  points  thu& 
found  will  indicate  the  curve. 

Let  R  be  a  point  found  as  above  described.  Join  FR,  and 
draw  RS  perpendicular  to  AB. 


By  construction 

MF 

CF  : 

:  m  \  n. 

But 

MF 

CF  : 

:  a.l=RF 

C.1=RS 

Therefore 

RF 

RS  : 

\  m  :  n. 

In  the  same  manner  it  may  be  shown,  that  the  distance  of  anv 


DEFINITIONS  AND  GENEBAL  PROPOSITIONS.  11 

other  point  in  the  curve  from  F  is  to  its  distance  from  AB  in  the 
given  ratio ;  and  hence  the  curve  is  a  conic  section. 

(3)  The  given  point  F  is  called  the  focus,  and  the  given  straight 
line  AB  the  directrix, 

(4)  The  portion  of  CX  intercepted  between  its  intersections  with 
the  curve,  is  called  the  transverse  or  major  axis. 

(5)  The  middle  point,  as  C,  (Fig.  2,)  of  the  transverse  axis  is 
called  the  centre  of  the  curve,  and  its  extremities,  A  and  B,  vertices, 

(6)  The  distances  from  the  focus  to  the  vertices,  FA  and  FB, 
are  called /oca?  distances. 

(7)  The  conjugate  axis,  DE,  is  a  straight  line  drawn  through  the 
centre,  at  right  angles  to  the  transverse  axis,  bisected  by  it,  and       7 
equal  to  twice  the  mean  proportional  between  the  focal  distances.       <^ 
Since  the  parabola  intersects  CX  in  only  one  point,  its  transverse 
axis  is  infinite,  and  it  has  no  conjugate  axis  nor  centre. 

(8)  Any  straight  line  drawn  through  the  centre,  and  limited  both 
ways  by  the  curve,  is  called  a  diameter,  and  its  extremities  its 
vertices ;  as  HI  (Fig.  6.) 

(9)  Two  diameters  are  said  to  be  conjugate,  when  each  is  parallel 
to  a  tangent  to  the  curve  at  the  extremity  of  the  other ;  as  NU  and 
PL  (Fig.  10.)  Two  hyperbolas  so  drawn  that  the  transverse  axis  of 
one  is  the  conjugate  axis  of  the  other,  and  vice  versa,  are  called 
conjugate  hyperbolas;  and  such  hyperbolas  only  have  conjugate 
diameters. 

(10)  An  ordinate  to  any  diameter  is  a  straight  line  parallel  to  a 
tangent  at  its  vertex,  and  limited  in  one  direction  by  the  curve,  and 
in  the  other  by  the  diameter;  as  RV  (Fig.  2,)  or  DZ  (Fig.  14.)  If 
produced,  so  as  to  be  limited  in  both  directions  by  the  curve,  it  is 
called  a  double  ordinate  ;  as  EU  (Fig.  2.) 

(11)  The  portions  into  which  an  ordinate  divides  a  diameter  are 
called  abscissas. 


12 


CONIC  SECTIONS. 


(12)  The  parameter  of  any  diameter  is  the  third  proportional  to  i 
and  its  conjugate.    In  the  parabola  it  is  the  third  proportional  to  any 
abscissa  and  its  corresponding  ordinate.    The  parameter  of  the  trans- 
verse axis  is  called  \\\&  principal  parameter,  or  latus-rectum. 

(13)  The  lines  KL  and  PQ  are  called  focal  tangents. 

(14)  The  distance  from  the  focus  to  the  centre  is  called  the  ec- 
centricity. 

(15)  A  line  drawn  perpendicular  to  a  tangent  to  the  curve,  from 
the  point  of  contact,  is  called  a  normal  line ;  and  the  part  of  the 
transverse  axis  intercepted  between  it  and  an  ordinate  let  fall  from 
the  point  of  contact,  is  called  a  subnormal. 

(15a)  Cor.  The  distance  from  any  point  in  the  curve  to  the 
focus  is  equal  to  a  perpendicular  to  the  transverse  axis,  drawn 
through  the  point  from  the  focal  tangent. 

(16)   Prop.  II.     Problem. 

To  describe  a  conic   section  that  shall  pass  through    three  given 
points,  and  have  a  given  focus. 

Let  M,  N,  and  P  be  the  three  ^  (Fig.  la.) 

given  points,  and  F  the  given 
focus.  We  have  now  to  find  the 
directrix. 

Join  FM,  FN,  FP,  MN,  and 
NP,  and  produce  MN  and  NP 
to  E  and  L,  making 

MR  :  NR  ::  MF  :  NF, 
and      NL  :  PL   ::  NF  :  PF. 

Through  the  points  R  and  L 
draw  the  line  ST  of  indefinite 
length,  and  it  will  be  the  re- 
quired directrix. 


DEFINITIONS  AND  GENERAL  PROPOSITIONS.  13 

For,  let  fall  upon  it  the  perpendiculars  PI,  NH,  MG,  and  FK. 

By  sim.  tri.  GM  :  HN  ::  MR  :  NR. 

But,  by  construction,  MF  :  NF    : :  MR  :  NR.  v 

Therefore,  alternately,  GM  :  MF   ::  HN   :  NF. 

Also  sim.  tri  IP     •  HN  ::  PL    •  NL. 

And  by  construction,  PF    :  NF    : .  PL    :  NL. 

Therefore,  alternately,  IP     :  PF    ::  HN  :  NF. 

Thus  the  distance  of  each  of  the  points  M,  N,  and  P,  from  the 
line  ST,  is  to  its  distance  from  F  in  the  same  ratio ;  and,  conse- 
quently, if  with  this  ratio  we  describe  a  conic  section,  in  the  same 
manner  as  in  Prop.  I.,  making  F  the  focus  and  ST  the  directrix,  it 
will  pass  through  the  points  M,  N,  and  P.  ^ 

This  proposition  is  important,  as  it  enables  us  to  determine  the       ^*- 
orbit  of  a  planet  or  comet  by  means  of  three  observations."^  ^ 

(16a)  Scholium.  The  conic  sections  may  be  formed  by  the  in- 
tersection  of  a  plane  with  the  sides  of  a  cone,  and  hence  their 
name.  If  the  cutting  plane  be  parallel  to  one  of  the  sides  of  the 
cone,  the  curve  is  a  parabola ;  if  more  nearly  perpendicular  to  the 
axis  of  the  cone,  it  is  an  ellipse ;  and  if  less  so,  an  hyperbola.  W 
quite  perpendicular,  the  section  is  evidently  a  circle.  And,  univer- 
sally, the  ratio  mentioned  in  (1)  is  the  ratio  of  the  sines  of  the 
angles,  which  the  cutting  plane  and  the  sides  of  the  cone  form  with 
the  base.     See  Appendix,  Note  A. 

•  Bridge's  Conic  Sections. 


14 


CONIC  SECTIONS. 


CHAPTER  II. 

OF    THE    ELLIPSE. 


(17)  Prop.  I.     Theorem. 

The  squares  of  ordinates  to  the  transverse  axis  of  an  ellipse  are  to 
each  other  as  the  'rectangles  of  the  corresponding  abscissas. 

That  is,     GP  :  RV2  ::  AF.FB  :  AV.VB. 

In  which  GF   is   the  focal  ordinate,  and  RV  any  other  ordi- 
nate. 

Through  the  vertices  A  and  (Fig.  2.) 

B  draw,  from  the  focal  tangent 
IT,  the  hnes  LM  and  IK  at 
right  angles  to  AB;  through 
the  focus  F  draw  LF  and  IF, 
which  when  produced  will 
meet  LM  and  IK  in  M  and  K ; 
and  produce  RV  both  ways  to 
P  and  N". 

By  the  principles   of  con- 
struction (15*), 

IB=BF,  and 

But*    IB  :  BF  ::  SV  :  FV,  and  AL  :  AF  ::  VN 

Therefore     SV  =  FV,  and  VN=FV. 
By  similar  triangles  (LGrF  and  LPN)  we  have 

GF  :  PN  ::  LG  :  LP  ::  ''AF  :  AV, 
and  by  similar  triangles  (GIF  and  IPS)  we  have 

GF  :  PS  ::  GI  :  PI  ::  "FB  :  VB. 

»  The  following  references  are  to  Loomis's  Geometry,  revised  edition,  and 
to  Wentworth's  Geometry.     L.,  4,  19;  W.,  3,  4,  and  5. 
»>  L.,  4,  16  ;  W.,  3,  2. 

'  Leg.  4.  18.     Euc.  6.  4.  "  Leg.  4.  15,  Cor.  2.     Euc.  6.  2. 


FV. 


OF  THE  ELLIPSE. 


15 


Multiplying  the  proportions  together, 

GP  :  PN.PS  ::  AF.FB  :  AV.VB. 

But     PS=PV-SV=(by  2)  FR-SV=FR-FV; 
and     PN=PV+VN=(by  2)  FR+VN=FR+FV. 


Therefore     PN.PS=FR+FV.  FR-FV=:'FRa-FVa=»»RV». 

Substituting  this  value  of  PN.PS,  we  have 

GP  :  RV«  ::  AF.FB  :  AV.VB. 

(18)  Cor.  1.  Hence,  if  tw^o  ordinates  are  equally  distant  from 
tftC  centre,  they  are  equal  to  one  another. 

That  is,     if  CF=CV,  then  GF=RV. '' 

(19)  Cor.  2.  Hence,  the  two  portions  of  the  curve  lying  on  the 
opposite  sides  of  the  transverse  axis  AB,  or  the  conjugate  axis  DE, 
are  symmetrical ;  and  if  placed  upon  one  another,  would  coincide 
in  every  part.  For  if  at  any  point  they  should  not  coincide,  the 
ordinates  at  that  point  would  be  unequal,  which  by  Cor.  1  is  im- 


(20)  Cor.  3.  Hence,  there  is  another  point  situated,  in  respect 
to  the  curve,  precisely  like  the  focus  F,  and  may,  therefore,  be 
called  another  focus.  Thus,  if  CV=CF,  .he  point  V  is  the  other 
focus. 


(21)  Cor.  4.  If  different  ellipses 
have  the  same  transverse  axis,  the 
corresponding  ordinates  are  propor- 
:ional  to  each  other. 

That  is,     FH  :  FH'  ::  GS  :  GS'. 


(Pig.  3.) 


Leg.  4.  10.     Enc.  2.  5,  Cor. 
L,  4. 10;  W.,§377. 


'  Leg.  4.  11.     Euc.  1.  47. 

^  L.,  4,  11;  W.,4,8. 


16 


CONIC  SECTIONS. 

For  (17) 

FH^ 

:  GS2    ::  AF.FB  :  AG.GB. 

Also, 

FH'2 

:  GS'2  ::  AF.FB  :  AG.GB. 

Therefore 

FH2 

:FH'2::GS2        :  GS^^. 

And-^ 

FH 

:  FH'  ::  GS          :  GS'. 

(22)  Cor.  5.     It  follows  from  the  last  of  the  foregoing  propoi 
tions,^  that  HH'  :  SS'  ::  FH  :  GS. 

(23)  Cor.  6.  Since  OC  is  midway  between  AL  and  BI,  it  equals 
half  their  sum.  But  BI+AL=BF+AF=AB.  Therefore  OC,  oi 
its  equal  FD,  is  equal  to  AC,  the  semi-transverse  axis. 


(24)  Prop.  II.     Theorem. 

The  square  of  an  ordinate  to  the  transverse  axis,  is  to  the  rectangle 
of  the  corresponding  abscissas,  as  the  square  of  the  conjugate  axis 
is  to  the  square  of  the  transverse  axis. 

That  is  (Fig.  2),     RV^  :  AV.VB  ::  DE^  :  AB« 

For  by  similar  triangles  (IGF  and  ILM)  we  have 

GF  :  LM=2AF  ::  IG  :  IL  ::  FB  :  AB. 

And  by  similar  triangles  (LGF  and  LIK)  we  have 
GF  :  IK=2FB  ::   LG  :   IL  ::  AF  :  AB. 

Multiplying  the  proportions  together, 

GP  :  4AF.FB=(7)  DE^  ::  AF.FB  :  AB«. 

But  (17)  GF2  :  RVa  ::  AF.FB  :  AV.VB. 

Therefore,  by  equality  of  antecedents,'^ 

RV2  :  AV.VB  ::  DE^  :  ABl 

(24a)  Cor.  1.     The  line  GH  is  the  parameter  of  the  transverse 
axis. 
For,  as  above,     GF^  ;  DE^  ::  AF.FB=JDE2  .  ^Bl 

•  Leg.  2.  12,  Cor.  *  Leg.  2.  6.    Euc.  6,  D.  and  16 

•  Leg.  2.  4.    Euc.  6.  24 

•  L.,  2, 11 ;  W.,  3, 10.  ^  L.,  2,  6  and  7 ;  W.,  3,  6  and  7. 

•  L.,  2,  4,  Cor. 


\ 

OF  THE  ELLIPSE.  .  I7 

Therefore,  extracting  roots,* 

AB  :  iDE  ::  DE  :  GF=iGH 
Or,  doubling  the  second  and  fourth  terms, 

AB  :    DE    ::  DE  :  GH  =  the  parameter,  which 
we  shall  hereafter  designate  by  the  letter  p. 

(25)  Cor.  2.  If  a  circle  be  described  on  the  transverse  axis  of 
an  ellipse,  an  ordinate  to  the  ellipse  is  to  the  corresponding  ordinate 
to  the  circle,  as  the  conjugate  axis  is  to  the  transverse. 

For  (Fig.  3) 

GS2  :  AG.GB=''GS'"2  ;:  DE«  :  AB«. 

Hence,  extracting  roots, 

GS  :  GS'"  ::  DE  :  AB. 

(26)  Cor.  3.  If  the  conjugate  axis  of  an  elHpse  is  equal  to  the 
transverse,  the  ellipse  becomes  a  circle.  For  then  the  square  of 
the  ordinate  becomes  equal  to  the  rectangle  of  the  corresponding 
abscissas,  which  is  a  known  property  of  the  circle.'* 

(27)  Cor.  4.  The  extremities  of  the  conjugate  axis  of  an  ellipse 
lie  in  the  curve.     For 

CD«=FDa-FC2=(23)  AC«-FC2=«AF.FB. 
which,  by  (7),  is  equal  to  the  square  of  the  semi-conjugate  axis 

(28)  Prop.  III.     Theorem. 

The  sum  of  two  line,%  drawn  from  the  foci  of  an  ellipse  to  any  point 
in  the  curve,  is  equal  to  the  transverse  axis. 

That  is,    VM+FM=AB. 

Make  the  arc  AN  equal  to  BM,  join  FN,  draw  the  ordinates 
NR  and  MS,  the  semi-conjugate  axis  DC,  and  the  focal  tangent 
TI,  and  produce  NR,  MS,  and  CD  to  G,  P,  and  O. 

•  Leg.  2.  12,  Cor.  *  Leg.  4.  23,  Cor.     Euc.  6.  8,  Cor. 

«  Leg.  4.  10.     Euc.  2.  6,  Cor. 

«  L.,  2, 11 ;  W.,  3. 10.  "  L.,  4,  23,  Cor. ;  W.,  2, 14,  Cor.  1. 

<=  L.,  4,10;  W.,§377. 

3 


19 


CONIC  SECTIONS. 
(Fig.  4.) 


Then,  since  CO  is  nmidway  between  GR  and  PS,  it  is  equal  ta 
naif  their  sum.     Therefore 

GR+PS=20C=(23)  AB. 

But  (2)      GR+PS=FN+FM=(19)  VM+FM. 

Therefore  VM+FM=AB. 

(29)  Scholium.  The  property  proved  in  this  proposition  lur- 
nishes  the  definition  of  the  ellipse  in  many  treatises.  It  also  afford* 
a  ready  method  of  describing  the  curve  mechanically.  Take  a 
thread  of  the  length  of  the  transverse  axis,  and  fasten  one  of  the 
ends  to  each  of  the  foci.  Then  carry  a  pencil  round  by  the  thread, 
keeping  it  alv^^ays  stretched,  and  its  point  will  describe  the  ellipse. 
For  in  every  position  of  the  pencil,  the  sum  of  the  distances  to  the 
foci  will  be  equal  to  the  entire  length  of  the  string. 


(30)  Prop.  IV.     Theorem. 

Two  lines  drawn  from  the  foci  to  any  point  in  the  curve,  make  equal 
angles  with  a  tangent  to  the  curve  at  that  point. 

That  is,  FM  and  VM  make  equal  angles  with  the  tangent  TU 
or  FMT=VMU. 


OF  THE  ELLIPSE. 


19 


If    not    let    them   make  (Fig*  *•) 

equal    angle^  with  RS,  so 
that  FMR=VMS. 

Since  there  cannot  be 
two  different  tangents  to  a 
curve  at  the  same  point, 
RS  must  cut  it.  Let  it 
cut  in  M  and  E. 

Produce  FM  to  G,  ma- 
king MG  =  MV,  and  join 
GV,  GE,  EF,  and  EV. 

The  angle  GMS=RMF=  (by  supposition)  VMS.  Then,  in  the 
triangles  GMD  and  VMD,  GM=MV,  and  MD  is  common,  and 
the  angle  GMD=VMD ;  therefore  GD= YD,  and  the  angle  GDM 
-VDM.  Hence,  in  the  triangles  EGD  and  EVD,  the  sides  GD 
and  ED=VD  and  ED,  and  the  angle  GDE=VDE;  therefore 
GE=EV,  and  EF+GE=EF+EV=(28)FM+MV=FG;  that  is, 
two  sides  of  a  triangle  are  equal  to  the  third  side,  which  is  impos- 
sible.* In  the  same  manner  it  may  be  shown,  that  no  other  line 
but  TU  makes  equal  angles  with  FM  and  MV,  and  consequently 
TU  does. 

(31)  Cor.  1.     Hence,  to  draw  a  tangent  to  the  curve  at  any 
point  M,  join  MF  and  MV,  and  bisect  the  exterior  angle  VMG. 

(32)  Cor.  2.     GV  is  perpendicular  to  TU,  and  is  bisected  at  the 
point  L 

(33)  Cor.  3.     FG=AB,  since  each  is  equal  to  FM+MV. 


•  Leg.  1.  7.    Euc.  1.  20. 

•  L.,1,  19;  W.,1,  26. 


eo 


CUi\IC  SECTIONS. 


(34)  Prop.  V.     Theorem. 

If  a  line  he  drawn  from  either  focus  perpendicular  to  a  tangent  to 
the  curve  at  any  point,  the  distance  of  its  intersection  from  the 
centre  is  equal  to  the  semi-transverse  axis.  ^ 


That  is,     CL=AC. 

Since  FV  is  bisected 
m  C,  and  (32)  GV  in  L, 
CL  is  parallel*  to  FG,  and 
the  triangles  LCV  and 
GFV  are  similar.    Hence 

CV  :  FV  ::  CL  :  GF. 


But  CV=LFV, 


(Fig.  6.) 


Therefore    CL=iGF=(33)  |AB=AC. 

(35)  Cor.  1.  Hence,  a  circle  described  on  the  transverse  axis 
with  the  centre  C,  will  pass  through  the  intersecfions  L  and  P ; 
and  conversely,  if  from  any  point  in  the  circumference  of  such  a 
circle,  two  lines  be  drawn  at  right  angles  to  one  another ;  and  if 
one  of  them  pass  through  one  of  the  foci,  the  other  will  be  tangent 
to  the  curve. 

(36)  Cor.  2.  Hence,  a  diameter  HI,  parallel  to  TU,  would  cut 
off  a  part  MX  of  the  line  MF,  equal  to  the  semi-transverse  axis 

For^    MX=CL=AC 


(37)   Cor.  3.     Since  CL   is   parallel   to  FM,  the  angle  CLM: 
FMP=(30)VML. 


•  Leg  4.  16.     Euc.  6.  2. 
«  L,  4, 16,  Cor. ;  W.,  3,  3. 


"  r.eg.  I.  28.     Euc.  1.  34. 
»»  L.,  1,30;  W.,  1,39. 


OF  THE  ELLIPSE. 


21 


(38)  Prop.  VI.     Theorem. 

The  rectangle  of  the  two  perpendiculars  drawn  from  the  foci  to  any 
tangent  to  the  curve,  is  equal  to  the  square  of  the  semi-conjugate 
axis. 

Thai  is,     VL.FP=CD2. 


Join  PC,  and  produce 
PC  and  LV  till  they  meet 
inN. 


(Fig.  7.) 


Then  will  the  triangles 
PFC  and  NVC  be  simi- 
lar and  equal,  since  the 
angle  PCF  =  NCV,  and  ' 
FPC='^the  alternate  angle 
VNC,  and  the  side  FC= 
VC.  Therefore  CN=CP, 
and  (35)  the  point  N  is 
in  the  circumference  of  a 
circle  described  on  A B  as 
a  diameter.  Consequently,^  NV.VL=AV.VB 
PF,  PF.LV=A^  VB=(7)CD2. 


Or,  since  NV=- 


(38a)  Prop.  VII.     Theorem. 

If  at  any  point  in  the  curve  a  tangent  and  ordinate  he  drawn,  meet- 
ing either  axis  produced,  half  that  axis  is  the  mean  proportional 
between  the  distances  of  the  two  intersections  from  the  centre. 


That  is, 

CS 

CA  : 

:  CA 

CT. 

Or, 

CG 

:  CD  : 

:  CD 

:  CH. 

Leg.  L  20,  Cor.  2.     Euc.  L  29 
L..1.  22:  W..  1. 13  and  14. 


"  Leg.  4.  28,  Cor.    Euc.  3.  36. 
b  L.,  4,  28,  Cor.;  W,  3,  11. 


22 


CONIC  SECTIONS. 

(Fig.  8.) 


Connect  M  with  the  foci  F  and  V,  draw  VL  perpendicular  to 
the  tangent,  and  join  CL  and  SL.    , 

Since  M6V  and  MLV  are  both  right  angles,  each  is  an  angle  in 
a  semicircle,*  and  consequently,  a  circle  described  on  MV  as  a 
diameter,  would  pass  through  L  and  S.  Then  must  the  angles 
VML  and  VSL  be  equal,  being  in  the  same  segment,"  or  measured 
by  the  same  arc. 

But  (37)  VML  =  CLM.  Therefore  the  angles  VSL  and  CLM 
are  equal,  as  also  their  supplements  CSL  and  CLT.  Hence,  the 
triangles  LCT  and  SCL  are  similar,  for  the  angle  CSL  =  CLT,  and 
the  angle  at  C  is  common. 

Therefore     CS  :  CL  ::  CL  :  CT. 
But  (34)  CL=CA. 

Therefore     CS  :  CA  ::  CA  :  CT, 
which  proves  the  proposition  in  respect  to  the  transverse  axis. 

(SSb)  Again,  since  when  three  numbers  are  in  continued  pr  por- 
tion, the  first  is  to  the  third  as  the  square  of  either  antecedent  if  ♦.o 
the  square  of  its  consequent,  we  have  from  the  last  proportior 
CS^  :  CA2  ::  CS=GM  :  CT  ::  (sim.  tri.)  GH  :  CH. 

Hence,  by  division,"    CA«-CS«  :  CA^  ::  CG  :  CH. 


Leg.  3.  18,  Cor.  2.     Euc.  3.  31. 
Leg.  2.  6.     Euc.  5.  17. 
L.,  3,  15,  Cor.  3  ;  W.,  3, 14,  Cor. 
L.,  2,  6  and  7;  W.,  3,  6  and  7. 


*  Leg.  3.  18.     Euc.  3.  21. 

»>  L.,  3,  15,  Cor.  1 :  W.,  2,  14,  Cor.  4. 


OF  THE  ELLIPSE. 


23 


But  (24)       AS.SB=K:A2-CS2  •  CA^  ::  MS«=CG*  :  CD«. 
Therefore,  by  equality  of  ratios, 

CG  :  CH   ::  CG^  :  CD^; 
which  gives,    CG.CH=CD*. 

Or,     CG  :  CD    ::  CD    :  CH. 

(39)  Prop.  VIll.     Theorem. 

If  different  ellipses  have  the  same  transverse  axis,  the  corresponding 
sub-tangents  are  equal  to  one  another. 

Let  EBDA,  E'BD'A,  &c.  (Fig-  9.) 

be  any  number  of  ellipses  de- 
scribed on  AB  as  the  trans- 
verse axis,  SG  produced  any 
ordinate  to  each,  and  ST, 
S'T,  &c.  tangents  at  the 
points  where  the  ordinate 
meets  the  curves. 

Then  for  each  of  them  we 
have  (38fl) 

CG  :  CA  ::  CA  :  CT, 

and  since  the  first  three  terms  are  the  same  for  all,  tne  fourth  must 
be  likewise,  which,  diminished  by  CG,  gives  the  sub-tangent  TG. 

(39a)  Cor.  1.  Hence,  we  may  draw  a  tangent  at  a  given  point 
in  the  curve,  without  knowing  the  foci. 

Let  S  be  the  given  point.  On  AB  describe  a  circle ;  draw  the 
ordinate  SG,  and  produce  it  till  it  meets  the  circle  in  S".  Draw* 
S"T  tangent  to  the  circle  at  S",  and  join  TS. 

(S9h)  Cor.  2.     CG.GT=AG.GB,  each  being  equal«  to  S"G«. 


'  Leg.  4.  10.    Euc.  2.  5,  Cor.  "  Leg.  3,  Prob.  14.    Euc.  3.  17. 

•  Lfig.  4.  23.     Euc.  6.  8,  Cor.  and  13. 

»  L.,  4,  10  ;  W.,  §  377.  ^  L.,  5,  Prob.  14  ;  W.,  2,  39. 

«  L.,  4,  23;  W.,  2,  14. 


24 


CONIC  SECTIONS. 


(40)  Prop.  IX.     Theorem. 

The  sum  of  the  squares  of  two  ordinates^  drawn  to  either  axisfrwn 
the  ext/remities  of  any  two  conjugate  diameters^  is  equal  to  the 
squa/re  of  half  the  other  axis. 

That  is,    FU2+GP2=AC^  and  P8^+UR-^=CEl 

Draw  the   tangents  PT  (Fig.  lo.) 

and  UK,  meeting  the  trans- 
verse axis  in  T  and  Y,  and 
the  conjugate  axis  in  H 
andK.    Then 

CS.CT=CR.CY, 

each  being  equal  (38a)  to 
AC^orBC^; 

Therefore*    OS  ;  OR  ::  CV  :  CT. 

But,  since  (9)  UY  is  parallel  to  PC,  and  UC  to  PT,  the  triangles 
PTC  and  UCY  are  similar ;  as  also  PCS  and  UYE. 

Hence,         YE  :  CS  : :  UY  :  PC  ::  CY  :  CT. 

Then,  by  equality  of  ratios,  we  have 

CS  :  CR  ::  YK  :  CS; 
And  consequently    CR.YR=CSl 

But  (395)      CR  .YR=AR.BR=^AC2- CRl 
Therefore     AC2-CR2=CS2;  or,  CR2+CS2=AO». 
That  is,        FU2+GP=AC2. 


•  Leg.  2.  2.    Loomis  2.  2.    Euc.  6.  16. 

"  Lee.  4.  10.    Loomis  4. 10     Euo.  2.  5,  Oor. 

•  W.,  3,8. 

b  W.,  §377. 


OF  THE  ELLIPSE. 


25 


Again,  by  comparing  the  similar  triangles  UKC  and  PCH,  and 
also  UCF  and  PHG,  it  may  be  proved  in  the  same  manner  that 

PS2+UK=^=CEl 

(41)  Gov,    AE.  BK=CS2,  each  being  eqnal  to  GR.VE. 

(42)  Prop.  X.     Theorem. 

The  sum  of  the  squares  of  any  pair  of  conjugate  diameters  is  equal 
to  the  sum  of  the  squares  of  the  two  axes. 

That  is,     UNa+PLa=AB2+DE«. 
(Fig.  11.) 


For»  UC«+PC2=CR2+CS2+UR2-fPS«. 

But  (40)  CR2+CS2=AC^  and  PS8+UR8=CE«. 

Therefore  UC2+PC2=AC3+CEl 

Or,  UN2+PL2=AB'»+DE^. 

(43)  Prop.  XI.     Theorem. 

Any  parallelogram  circumscribed  about  an  ellipse^  having  its  sides 
parallel  to  two  conjugate  diameters,  is  equal  to  the  rectangle  of  the 
two  axes. 

That  is,    GHIK=LMNO=ABxDE. 


•Leg.  4.  11.    Euc.  1.  47. 
-  L.,  4, 11  ;  W.,  4,  8. 

4 


26 


CONIC  SECTIONS. 
(Fig.  12.) 


Draw  CX  at  right  angles  to  TK ;  then  will  the  triangles  TCX 
and  CUR  be  similar. 

mw  (24)    DC^ :  AC^ : :  UR^ :  AE.RB=  (41)  CS^ 

Therefore,  extracting  roots, 

DC  :  AC  ::  UR  :  CS. 

Or,  alternately,    AC  :  CS  ::  DC  :  UR. 

But  (38a)  AC  :  CS   ::  CT  :  AC. 

Therefore  DC  :  UR  ::  CT  :  AC, 

or*  UR.CT=DC.AC=iAB.DE. 

Also  CU  :  UR  ::  CT  :  CX, 

or   UR.CT=CU.CX=HGHIK. 

Therefore  GriIK=AB.DE. 

(44)  Cor,  1.    AC  :  CX  ::  CU  :  DC,     or   CU.CX=AC.DC. 


•  Leg.  2.  2.    Euc.  6.  16. 
•L.,2,2;  W.,3,  3. 


•  Leg.  4.  6. 
t>L.,4,  5;  W.,4,4. 


or  THE  ELLIPSE. 


37 


(45)  Prop.  XII.     Theorem. 

The  rectangle  of  two  lines,  drawn  from  the  foci  of  an  ellipse  to  any 
point  in  the  curve,  is  equal  to  the  square  of  half  the  diameter 
parallel  to  the  tangent  at  that  point. 

That  is,     FM.VM=CHl 

Draw  the  tangent  LMP,  and  the  (Fig-  13.) 

perpendiculars  to  it  FL,  MN,  and 
VP.  Then  will  the  triangles  PMV, 
SMN,  and  FML  be  similar  (30). 

Therefore 

MS  :  MN  ::  FM   :  FL. 
And  also, 

MS  :  MN  ::  VM  :  VP. 

Multiplying  the  proportions  together, 

MS2  :  MN2  ::  FM.VM  :  FL.VP. 

But  (36)         MS^r-AC^,      and  (38)   FL.VP=rCE«. 

Therefore        AC^  :  MN^  ::  FM.VM  :  CE«. 

Now  (44)  MN.CH=AC.CE. 

Therefore        AC    :  MN    ::  CH    :  CE. 

Or,  squaring,  AC  :  MN^  ::  CH^  :  CEl 

Hence,  by  equality  of  ratios,    CH^  :  CE^  ::  FM.VM  :  CBF. 

And,  consequently,        FM.VM=CH2. 


•  Leg.  4.  6. 

•L..  4.5;  W.,4  4. 


28 


COiNIC  SECTIONS. 


(46)  Prop.  XIII.     Theorem. 

If  at  one  of  the  vertices  of  an  ellipse  a  tangent  be  drawn  meeting 
any  diameter  produced,  and  also  from  the  same  point  an  ordinate 
to  that  diameter,  the  semi-diameter  is  the  mean  proportional  fee- 
tween  the  distances  of  the  two  intersections  from  the  centre. 

That  is,     CE  :  CI  ::  CI  :  CP. 

(Fig.  14.) 


Draw  IT  and  10  tangent  and  ordinate  to  the  curve  at  I.     Theiu 

by  similar  triangles, 

CE  :  CI    ::  CB  :  CT. 


CO  :  CB  ::  CB  :  CT. 


But  (38a) 

And,  by  similar  triangles, 

CO  :  CB  ::  CI    :  CP. 

Hence,  by  equality  of  ratios, 

CE  :  CI    ::  CI    :  CP. 


(47)  Cor.  1.     Hence,  CP  :  CP»::  CE  :  CP;  and,  in  like  man- 
ner.  CN^  :  CR^  ::  CS  :  CR. 


(48)  Cor.  2.  The  lines  CP  and  CT  are  similarly  divided,  the 
former  in  the  points  I  and  E,  and  the  latter  in  B  and  O ;  and,  con- 
sequently, *lines  j'oining  EO,  IB,  and  PT,  would  be  parallel. 


"  Leg.  4.  16.     Enc.  6  2 
•L.,4,  16;  W.,3,  2. 


OF  THE  ELLIPSE.  29 

(49)  Cor.  3.  The  area  of  the  triangle  CIT=CBP,  and  CEBr^ 
COI ;  for  the  angle  at  C  is  common,  and  the  sides  about  it  recip- 
rocally proportional.*     In  like  manner,  CBR=CNF 

(60)  Cor,  4      The  area  of  the  triangle  CNG=CBP  or  CIT.* 
For**     CNG  :  CRP  ::  CN^  :  CR^  ::  .(Cor.  1)  CS=BE  :  CR  : 

(sim.  tri.)  BP  :  PR  ::  '^CBP  :  CRP. 

Hence,  since  CNG  and  CBP  have  the  same  ratio  to  CRP,  they 

are  equal  to  one  another. 

(51)  Cor.  5.     IOBP=IOT. 

For  (49)  CIT=CBP,  and  taking  CIO  from  each,  the  remainders 
must  be  equal. 

(52)  Cor.  6.  The  triangle  UZX=PBXL,  (Z  being  any  point  m 
the  curve.) 


For      CB  :  BP  ::  CX  :  XL 

Also,    CB  :  BP  ::  CO  :  01 
Therefore  AX  :  AO 


'^CB+eX=AX  :  BP+XL. 
CB+CO=AO   :  BP+OI. 
BP+XL  :  BP-fOI. 


Or,«      AX.XB  :  AO.OB  ::  BP+XL.XB  :  BP+OI.OB. 
But 


BP+XL.XB=^2PLXB,     and     BP+OI.OB=2PIOB=(51)  2I0T 

Therefore 
PLXB  :  lOT  :    AX.XB  :  AO.OB  ::  (17)  ZX^  :  OP  ::  ''UZX  :  lOT 

Hence,     PLXB=UZX,  since  both  have  the  same  ratio  to  lOT 

(53)   Cor.  7.     DZL=ICT-DUC. 

For  (49) 

ICT=CBP=CLX+PLXB=(52)CLX+UZX=DUC+DZL, 
since  the  part  ZDCX  is  common  to  both  UZX  and  DZL. 
Hence,     DZL= ICT  -  DUC. 


'  Leg.  4.  24,  Cor.     Euc.  6.  15.  "  Leg.  4.  25.     Euc.  6.  19. 

•  Leg.  4.  6,  Cor.     Euc.  6.  1.  ^  Leg.  2.  10.     Euc.  6.  12. 

'  Leg.  2.  8.     Euc.  6.  15  and  16.  '  Leg.  4.  7. 

»  L.,  4,  34 ;  W.,  4,  13.  »»  l.  4  35  .  w. ,  4.  14. 

c  L.,  4,  6,  Cor.  1 ;  W. .  4,  5,  Cor.  2.  «»  L.,  2,  9  ;  W.,  3,  8. 

«  L.,  2,  10  ;  W.,  3,  11.  '  L.,  4,  7;  W.,  4,  6. 
*  HI  and  MN  being  conjugate  dianietera 


30  CONIC  SECTIONS 

(54)   Prop.  XIV.     Theorem. 

The  square  of  any  diameter  is  to  the  square  of  iU  conjugate^  <m  the 
rectangle  of  its  abscissas  is  to  the  square  of  the  corresponding 
ordinate. 

That  is,     HP  :  MN^  ::  HD.DI  :  DZ?. 
(Fig.  16.) 


For*    CN^  :  DZ«  ::  CNG=-(50)  ICT  :  DZL=(53)  ICT-DUC 

But*  ICT  :  DUG  ::  CP  :  GDI 

Hence,"    IGT  :  IGT-DUG  ::  CP  :  GP-GD«=''HD.DI 

Therefore,  by  equality  of  ratios, 


CW  :  DZ2 
Or,  alternately,  GP  :  GN^ 
Or,  HP    :  MN2 


:  GP  :  HD.DI. 
:  HD.DI  :  DZl 
:  HD.DI  :  DZl 


(56)  Cor.  Hence,  all  chords  parallel  to  any  diameter  are  bi- 
sected by  its  conjugate ;  and,  conversely,  a  line  bisecting  two  or 
more  parallel  chords  is  a  diameter. 

(56)  Prop.  XV.  Problem. 

To  draw  a  tangent  to  an  ellipse  from  a  given  point  without  the 

curve. 


'  Leg.  4.  25.     Euc.  6.  19. 
•  Leg.  4.  10.     Euc.  2.  6,  Cor. 
»  L,  4,  25;  W.,4,  14. 
e  L.,4,  10;  W.,  §377. 


'  Leg.  2.  6.    Euc.  5,  D. 

b  L.,2,  6aiid7;W.,3,  6and7. 


OF  THE  ELLIPSE. 


31 


Let    AMB    be   the 

given  ellipse,  AB  the 
transverse  axis,  F  one 
of  the  foci,  and  T  the 
given  point. 

Join  TF,  and  upon 
It  and  AB  as  diame- 
ters, describe  the  cir- 
cles TPF  and  APB, 
cutting  each  other  in 
P  and  F.     The  lines 


(Pig.  16.) 


'Ube 


TPM  and  TP'M'  drawn  through  the  points  of  interseation 
tangents  to  the  ellipse. 

Join  FP.  The  angle  FPT  is*"  a  right  angle,  and  FP  perpendic- 
ular to  TM.  Now,  since  from  the  point  P,  in  the  circumference  oi 
the  circle  described  on  the  transverse  axis,  there  are  drawn  two 
lines,  PF  and  PM,  at  right  angles  to  one  another,  and  one  of  them 
(PF)  passes  through  the  focus,  the  other  must  (35)  be  tangent  to 
the  ellipse. 


(57)  Prop.  XVI.     Problem. 
To  find  tfie  centre,  axes,  and  foci  of  a  given  ellipse. 


Let  BDAE  be  the  given 
ellipse. 

Draw  any  two  pairs  of  par- 
allel chords  HI  and  LK,  MN 
and  OP;  bisect  them  in  T, 
U,  R,  and  S;  join  TU  and 
RS,  and  produce  the  lines  till 
they  meet  in  C.  Both  these 
lines  being  (55)  diameters, 
the  point  C  must  be  the 
centre. 


(Fig.  17.) 


•  Leg.  3.  18,  Cor.  2.     Euc.  3.  31 
«  L.,  3,  18,  Cor.  3 ;  W.,  2, 14,  Cor. 


32 


CONIC  SECTIONS. 


From  the  centre  C,  and  with  any  convenient  radius,  describe  a 
circle  cutting  the  ellipse  in  any  points  W,  X,  Y,  and  Z.  Draw  WX 
and  XY,  and  at  right  angles  to  them  respectively,  draw  AB  and 
DE  through  the  centre  C.  Since  these  lines  pass  through  the 
centre,  they  are  diameters ;  and  since  they  bisect  "^WX  and  XY  at 
right  angles,  they  divide  the  ellipse  into  two  similar  parts,  and 
therefore  are  (19)  ax6s. 

From  E,  the  extremity  of  the  conjugate  axis  as  a  centre,  and 
with  a  radius  equal  to  the  semi-transverse  axis,  describe  the  arc 
FQV,  cutting  the  transverse  axis  in  F  and  V.  These  points  ar*» 
(23)  the  foci. 


(57a)  Prop.  XVII.     Theorem. 

An  ellipse  may  be  formed  by  the  mutual  intersection  of  a  cone  ana 

a  plane. 

Suppose  the  ellipse  in  Prop.  I.,  with  no  change  of  letters,  to  be 
placed  upon  the  cone  CDLE  in  the  manner  of  a  collar,  with  its 
plane  perpendicular  to  the  triangular 
section  CDE,  the  latter  being  perpen- 
dicular to  the  base  of  the  cone,  and 
passing  through  A,  B,  and  C.  Now, 
suppose  the  point  A  to  slide  up  or  down 
on  the  line  CD,  and  B  on  CE,  till  the 
point  G  shall  lie  in  the  surface  of  the 
cone ;  a  condition  which  is  evidently 
possible,  whatever  be  the  nature  of  the 
curve  AGBH.  We  assert  that  then 
will  any  other  point  R  in  the  ellipse 
also  lie  in  the  surface  of  the  cone. 

If  not,  it  must  lie  either  within  or 
without  the  cone.  Let  it  be  supposed 
10   lie   without,  and   that   the   ordinate 


Leg.  3.  6.     Euc.  3.  3. 
L.,3,6;  W,  2,  7. 


or  THE  ELLIPSE.  33 

RV  cuts  the  surface  of  the  cone  at  s  Through  G  and  s  let  tne 
circular  sections  PGSH  and  MsNU  be  made  to  pass,  parallel  to 
the  base,  and  cutting  the  triangular  section  in  PS  and  MN.  The 
lines  GF  and  RV  being*  perpendicular  to  the  plane  CDE,  must  also 
be  perpendicular  to  PS  and  MN. 

By  sim.  tri  AFP  and  AVM,     PF  :  MV  ::  AF  :  AV. 

And  by  sim.  tri.  BFS  and  BVN,     FS  :  VN   ::  FB  :  VB. 
Multiplying  the  proportions  together, 

PF.FS  :  MV.VN  ::  AF.FB  :  AV.VB. 
But**  PF.FS=GF2,    and  MV.VN=*V«. 

Therefore         GP  :  sY^   ::  AF.FB  :  AV.VB. 
But  (17)  GF^  :  RV2  ::  AF.FB  :  AV.VB. 

Therefore        sV^=RV^     and  5V=RV,  which  is  impossible. 

In  the  same  manner  it  may  be  shown,  that  the  point  R  cannot  lie 
witnm  the  cone,  and,  consequently,  it  lies  in  the  surface.  And 
since  R  is  any  point  in  the  ellipse,  the  whole  curve  must  lie  in  the 
surface  of  the  cone. 

'  Leg.  6.  18.    Loomis  7.  8.    Euc.  Sup.  2.  18. 

*  Leg.  4.  23,  Oor.    Euc.  6. 18. 

•  W.,  6,  19. 

b  L.,  4,  33,  Cor. ;  W.,  2, 14,  Cor.  1. 
6 


34 


CONIC  SECTIONS. 


CHAPTER  III. 

OF    THE    PARABOLA 


(58)  Prop.  I.     Theorem. 

The  squares  of  ordinates  to  the  transverse  axis  of  a  paraboU  are 
to  each  other  as  the  corresponding  abscissas. 

That  is,     GF2  :  RV2  ::  AF  :  AV. 

Let  F  be  the  focus  (Fig.  19.) 

1 

p  / 

Draw  the  focal  tangent  IT,  and 
FS  parallel  to  it ;  from  the  vertex 
A  draw  AL  at  right  angles  to 
AX;  join  LF,  and  produce  RV 
both  ways  till  it  meets  IT  and 
LF  produced  in  P  and  N. 

Then,  because  (1)  FG  =  FT, 
and  (15^)  AL=AF,  we  have  also 
by  similar  triangles,  as  in  Prop.  I. 
of  the  ellipse,  AL=AF,  SV=FV, 

and  VN=FV. 

By  similar  triangles,  LGF  and  LPN,  we  have 

GF    :  PN  ::  LG  :  LP  ::  'AF  :  AV. 

Multiplying  the  first  couplet  by  GF=PS, 
GF2  :  PSxPN  ::    AF  :  AV. 

But,  as  in  Prop.  I.  of  the  ellipse,  PSxPN=RV«. 
Therefore,    GF^  :  RV^  ::  AF  :  AV 


•  Leg.  4.  15,  Cor.  2.    Euc.  6.  3. 
»  L.,  4, 15,  Cor.  3 ;  W.,  2,  2. 


OF  THE  PARABOLA.  35 

SchoL  If  we  suppose  the  parabola  to  have  anotner  vertex  at  an 
infinite  distance  in  the  direction  AX,  this  proposition  will  be  the 
same  as  Prop.  I.  of  the  ellipse. 

(59)  Cor.  1.  The  line  GH  is  the  parameter  of  the  transverse 
axis. 

For  since  (2)  in  the  parabola  the  angle  GTF=45°,  AT=AL 
=AF.     Hence  FT=2AF,  FG=2AF,  and  GH=4AF. 

Substituting  2AF  in  place  of  GF  in  the  last  proportion,  we  have 

4AF2  .  RV2  ::  AF    :  AV. 
Dividing  the  first  and  third  terms  by  AF,  we  obtain 

4AF    :  RV2  ::   1        :  AV. 
Or,     AV      :   1        ::  RV-^  :  4AF=GH. 
Or,  by  transferring  the  factor  RV, 

AV     :  RV    ::  RV    :  GH. 

(60)  Cor.  2.  By  the  third  proportion  in  the  foregoing  corollary 
we  learn  that  RV2=:AV'4AF=AV-GH  ;  that  is,  the  square  of  any 
ordinate  is  equal  to  the  abscissa  multiplied  by  four  times  the  focal 
distance,  or  by  the  parameter. 

(60«)  Cor.  3.  The  two  portions  of  the  curve  lying  on  the  oppo- 
site sides  of  the  transverse  axis  are  symmetrical. 

Props.  II.  and  HI.  of  the  ellipse  are  applicable  to  the  parabola 
only  upon  the  supposition  that  its  transverse  axis  is  infinite,  and 
that  it  has  another  focus  infinitely  distant. 


(61)  Prop.  IV.     Theorem. 

Two  lines  drawn  from  any  point  in  the  curve,  one  to  the  focus  and 
the  other  parallel  to  the  transverse  axis,  make  equal  angles  with  a 
tangent  to  the  curve  at  that  point. 

That  is,  FM  and  VM  make  equal  angles  with  TU,  or  FMT=:VMU. 


36 


'JONIC  SECTIONS. 


If  not,  let  them  make  equal  angles  with  RS,  so  that  RMF 
V^MS. 

Since  there  cannot  be  two 
different  tangents  to  a  curve  at 
the  same  point,  RS  must  cut  it 
and  fall  within,  as  at  some  point 
E.  Through  E  draw  the  ordi- 
nate KN  cutting  the  curve  in 
K.  Produce  VM  till  it  meets 
the  directrix  in  G ;  join  GF, 
GE,  EF,  and  KF,  and  draw 
KP  and  EO  parallel  to  AN. 

The  angle  GMR=:VMS  = 
(by  supposition)  RMF.  Then, 
in  the  triangles  GMD  and  FMD, 
GM=(1)  FM  and  MD  is  com- 
mon,   and    the    angle   GMD  = 

FMD ;  therefore  GD=FD,  and  the  angle  GDM=FDM.  Hence, 
in  the  triangles  EGD  and  EFD,  the  two  sides  GD  and  ED  are 
equal  to  FD  and  ED,  and  the  angle  GDE=FDE ;  therefore  EG=: 
EF.  Now  KF,  being  opposite  to  the  greater  angle  of  the  triangle 
EKF,  is  greater  "than  EF,  and  is  therefore  greater  than  EG.  But 
KF=(1)  KP=EO;  therefore  EO  is  greater  than  EG;  that  is,  one 
of  the  perpendicular  sides  of  a  right-angled  triangle  is  greater  than 
the  hypotenuse,  which  is  impossible. 

In  the  same  manner  it  may  be  shown  that  no  other  line  bui  TU 
makes  equal  angles  with  FM  and  VM,  and  consequently  TU  does. 

(62)  Cor.  I.  Hence,  to  draw  a  tangent  to  the  curve  at  any 
point  M,  join  MF,  draw  MG  parallel  to  the  transverse  axis,  and 
bisect  the  angle  FMG. 

(63)  Cor.  2.  GF  is  perpendicular  to  MT,  and  they  mutually 
bisect  each  other  at  the  point  H,  as  is  evident  from  the  equal 
triangles  GHM  and  PHT. 


'  Leg.  1.  13.    Euc.  1.  19. 

•L.,1,  13;  W.,  1,33  and  :}4. 


OF  THE  PARABOLA. 
(64)   Cor.  3.     FM=FT. 


87 


(65)  Cor.  4.     AN'=AT;  or  TN'=2AN';    that  is,  every  sub- 

tangent  is  bisected  at  the  vertex,  or  is  equal  to  twice  the  abscissa 
For  CN'=GM=FM=FT. 

And  (1)     CA  =AF. 
Therefore,  subtracting  equals, 

AN'=AT,  or  2AN'=TJN'. 

(66)  Cor.  5.  If  different  parabolas  have  the  same  transverse 
axis,  the  corresponding  subtangents  will  be  equal  to  one  another. 
For  in  each  case  the  sub-tangent  will  be  equal  to  twice  the  abscissa. 

(67)  Cor.  6.  Hence  we  may  draw  a  tangent  at  a  given  point, 
as  M,  in  the  curve,  without  knowing  the  focus  (62) ;  viz.,  draw  thp 
ordinate  MN',  make  AT  equal  to  AN',  and  join  TM. 


(68)  Prop.  V.     Theorem. 

If  a  line  he  drawn  from  the  focus  perpendicular  to  a  tangent  to  the 
curve  at  any  point,  a  tangent  at  the  vertex  will  pass  through  the. 
point  of  intersection. 

That  is,  if  FH  is  perpendicular  to  TM,  a  tangent  to  the  curve 
at  A  will  pass  through  H. 


Since  (63)  TM  is  bi- 
sected in  H  and  (65)  TN 
in  A,  AH  is  parallel  to 
MN. 

Hence  it  is  perpendic- 
ular to  AN,  and  conse- 
quently tangent  to  the 
curve  at  A.  And  as  there 
cannot  be  two  tangents  to 
a  curve  at  the  same  point, 
a  tangent  at  A  must  pass 
through  H 


(Fig.  21.) 


SH  CONIC  SECTIONS. 

(68a)  Cor.  1.  Hence,  if  a  line  be  drawn  from  F  to  any  poini 
H  in  ^le  tangent  at  the  vertex,  a  line  HM  drawn  from  H  perpen 
dicular  to  FH  will  touch  the  curve. 

(6Sh)  Cor.  2.  The  sub-normal  NG=2AF,  and  is  therefore  a 
constant  quantity,  wherever  in  the  curve  the  point  M  be  taken. 

For,  by  similar  triangles, 

TH  :  TM  ::  AH  :  MN  ::  AF  :  NG. 

But  (63)  TM=2TH,     therefore  NG=2AF. 

(68c)   Cor.  3.     FH2=AF.FM. 

For*^     AF  :  FH  ::  FH  :  FT=FM. 

{68d)   Cor,  4.     TM2=AN.4FM,    or  ^=4FM. 

For,  by  similar  triangles,     TF  :  TH  ::  TH  :  TA=AN. 
Therefore,  TH2=AN.TF=AN.FM. 

And,  consequently,     TM2=AN.4FM. 

Prop.  V.  of  the  ellipse  is  true  also  of  the  parabola,  if  we  regaru 
its  transverse  axis  as  infinite.  For  the  circumference  of  a  circle 
whose  diameter  is  infinite  is  a  straight  line,  so  that  AH  of  the  pre- 
ceding figure  may  be  considered  as  an  arc  of  a  circle  described  on 
the  transverse  axis. 

Now  it  has  been  shown  (68)  that  the  perpendicular  FH  meets 
TM  in  this  circumference,  and  therefore  at  a  distance  from  the 
centre  equal  to  the  radius,  that  is,  to  half  the  transverse  axis. 

Props.  VI.,  VII.,  IX.,  X.,  XI.,  and  XII.  of  the  ellipse  demonstrate 
properties  to  which  there  are  none  analogous  in  the  parabola. 

Pro?.  VIII.  of  the  ellipse  corresponds  to  (66). 


'  Leg.  1.  13.    Euc.  1.  19. 
•  L.,  1,13;  W.,  1,  33  and  34. 


OF  THE  PARABOLA. 


m 


(69)  Prop.  XIII.     Theorem. 

If  at  the  vertex  of  a  parabola  a  tangent  he  drawn  meeting  any 
diameter  produced,  and  also  from  the  same  point  an  ordinate  to 
that  diameter y  the  distances  of  the  intersections  from  the  cunw 
measured  on  the  diameter  will  he  equals 

That  is,     MP=ME. 

(Fig.  22.). 


For,  since  the  angle  PHM=AHT,  and  HPM=HAT,  and  (63 
and  68)  the  side  TH=HM,  we  have  AT=MP. 

But"  AT=ME. 

Therefore    MP=ME. 

(70)  Cor,  1.     The  triangle  HAT=HPM. 

(71)  Cor.  2.     The  lines  EP  and  OT  are  similarly  divided,  so 
that  lines  joining  EO,  MA,  and  PT,  would  be  parallel. 

(72)  Cor.  3.     The  triangle  OMT=APE=«PAOM=«»MTAE. 


■  A  line  drawn  parallel  to  the  transverse  axis  from  any  point  in  the  cnnre  la 
called  a  diameter. 

"  Leg.  1.  28.     Euc.  1.  34.  •  Leg.  4.  2.     Euc.  1.  41. 

"  Leg.  4.  1.     Euc.  1.  36. 

»•  L.,  1,  30 ;  W.,  1,  39.  •  L.,  4,  2  ;  W.,  4,  5. 

1  L.,  4,1;  W.,  4,  4,  Cor. 


iO 


CONIC  SECTIONS. 


(73)  Cor.  4.     The  triangle  UZX=PAXL,  (Z  being  any  point  m 
the  curve.) 

For  (58)     ZX2  :  OM^  ::  AX  :  AO  ::  "^PAXL  :  PAOM, 
But  (sim.  tri.  UZX  and  OMT)  we  have 

*>ZX2  :  0M2  ::  UZX  :  OMT. 
Therefore,  PAXL  :  PAOM  ::  UZX  :  OMT. 
But  (72)  OMT=PAOM ;   therefore  UZX=PAXL. 

(74)  Cor,  5.     DZL=DUAP=DUTM. 

For  DZL=DUXL+UZX=(73)  DUXL+PAXL=DUAP. 

(75)  Prop.  XIV.     Theorem. 

The  squares  of  ordinates  to  any  diameter  of  a  parabola  are  to  eacik 
other  as  the  corresponding  abscissas. 

That  is,    EA«  :  DZ^  ::  ME  :  MD. 

(Fig.  23.) 


p 

L          M/-                    E                                           1) 

H 

^ 

w 

^^ 

*^                 A 

X 

F    I 

'^ 

Foi    «EA«  :  DZ«  ::  APE=(72)  MTAE  :  DZL= 
(74)DUTM  ::  "^ME  :  MD. 


•Leg.  4.  3.    Euc.  6.  1. 
•  Leg.  4.  26.    Euc.  6.  19. 
a  L.,4,  3;  W,  4, 1. 
«  L.,4,  35;  W.,4, 14, 


'  Leg.  4.  26.    Euc  6.  19. 
b  L.,4,  25;  W.,4, 14. 


OF  THE   PAEABOLA. 


41 


,(76)  Cor,  1.  Hence  all  chords  parallel  to  a  tangent  at  any  point 
of  a  parabola  are  bisected  by  a  diameter  terminating  at  that  point ; 
and  conyersely,  a  line  bisecting  two  or  more  parallel  chords  is  a 
diameter. 

(76a)  Cor,  2.     ME  :  AE  ::  AE  :  4FM. 
For  (68^)    TM2=A0.4FM. 
But  TM=:AE,     and    AO=ME. 

Hence  AE2=ME.4FM. 

(765)  Cor.  3.     MD  :  DZ  ::  DZ  :  4FM ;    or    DZ2  =  MD.4FM, 

Z  being  any  point  in  the  curve. 


(7Y)  Prop.  XV.     Problem. 

To  draw  a  tangent  to  a  parabola  from  a  given  point  without  t/ie 

curve. 

Let    M'AM    be    the  (Fig.  24.) 

given  parabola,  AX  the 
transverse  axis,  F  the 
focus,  A  the  vertex, 
and  T  the  given  point. 
Join  TF,  and  upon  it 
as  a  diameter  describe 
the  circle  TPF,  cut- 
ting PAP',  the  tangent 
at  the  vertex,  in  the 
points  P  and  P'.  The 
Hues  TPM  and  TFM' 

drawn  through  the  points  of  intersection  will  be  tangents  to  the 
curve. 

Join  FP.  The  angle  TPF  is*  a  right  angle,  and  FP  perpendicular 
to  PM.  Then,  since  from  a  point  P  in  the  tangent  AP,  a  line  PM 
is  drawn  perpendicular  to  FP,  it  is  (68a)  a  tangent  to  the  curve. 


'  Leg.  3.  18,  Cor.  2.     Euc.  3.  31. 
»  L.,  3,  15,  Cor.  3 ;  W.,  2, 14,  Cor. 


42 


CONIC  SECTIONS. 


(78)  Prop.  XVI.     Problem. 
To  find  the  axis  and  focus  of  a  parabola. 


(Fig.  26.) 


Let  GEAD  be  the  given  parabola. 

Draw  any  two  parallel  chords 
DE  and  BC,  bisect  them  in  L 
and  N,  and  through  the  points 
of  bisection  draw  the  line  MU. 
Then  (76)  will  MU  be  a  diam- 
eter. Draw  DG  at  right  angles 
to  MU,  bisect  it  in  P,  and  through 
P  draw  AX  at  right  angles  to 
DG.  The  Hne  AX  is  the  trans- 
verse axis  (60a),  since  it  bisects 
the  chord  DG  at  right  angles. 

Through  M  draw  MH  parallel 
to  DE,  through  A  draw  AH  per- 
pendicular to  AX,  and  from  H,  the  point  of  their  intersection,  draw 
HF  perpendicular  to  HM.  Then,  since  AH  a  tangent  at  the  vertex, 
MH  a  tangent  at  the  point  M,  and  HF  a  perpendicular  to  the  latter, 
intersect  each  other  in  the  same  point  H,  the  point  F  must  (68a)  be 
the  focus. 


(78a)  Prop.  XVII.     Theorem. 

A  parabola  may  be  formed  by  the  mutual  intersection  of  a  cone  and 

a  plane. 

Suppose  the  parabola  in  Prop.  I.,  with  no 
change  of  letters,  to  be  placed  upon  the  cone 
CDLE  in  the  manner  of  a  collar,  with  its 
plane  perpendicular  to  the  triangular  section 
CDE,  and  parallel  to  the  side  CE,  the  latter 
section  being  perpendicular  to  the  base  of 
the  cone  and  passing  through  C  and  A. 

Now,  suppose  the  plane  RGAH  to  move 
parallel  to  itself,  yet  so  as  to  keep  the  point 


OF  THE  PARABOLA.  4H 

A  on  the  line  CD  till  the  point  G  shall  lie  in  the  surface  of  the 
cone,  a  condition  which  is  evidently  possible,  whatever  be  the 
nature  of  the  curve  RGAH.  We  assert  that  then  will  any  other 
point  R  in  the  parabola  also  lie  in  the  surface  of  the  cone. 

If  not,  it  must  lie  either  within  or  without  the  cone.  Let  it  be 
supposed  to  lie  without,  and  that  the  ordinate  RV  cuts  the  surface 
of  the  cone  at  s.  Through  G  and  s  let  the  circular  sections  PSGH 
and  DUE^  be  made  to  pass,  parallel  to  the  base,  and  cutting  the 
triangular  section  in  PS  and  DE.  The  lines  GF  and  RV  being 
perpendicular  to  the  plane  CDE,  must  also  be  perpendicular  to  PS 
and  ED. 

By  sim.  tri.,  AFS  and  AVD, 

AF  :  AV  ::  FS  :  VD. 
Multiplying  the  last  couplet  by  PF=*EV, 

AF  :  AV  ::  PF.FS  :  EV.VD. 
But''  PF.FS=FG2,    and  EV.VD=sV«. 

Therefore  AF  :  AV  ::  GP  :  sY^. 

But  (58)  AF  :  AV  : :  GP  :  RVl 

Therefore  5V^=RV2,     and  5V=RV,  which  is  impossible. 

In  the  same  manner  it  may  be  shown  that  the  point  R  cannot  lie 
within  the  cone,  and  consequently  il  lies  in  the  surface.  And  since 
R  is  any  point  in  the  parabola,  the  whole  curve  must  lie  in  the 
surface  of  the  cone. 


•  Leg.  1.  28.    Euc.  1.  34.  "  Leg.  4.  23,  Cor.    Euc.  6.  13. 

-  L.,  1,  30. ;  W.,  1,  39.  »»  L.,  4,  23,  Cor. ;  W.,  2, 14,  Cor.  1. 


44 


CONIC  SECTIONS. 


CHAPTER  IV. 

OF    THE    HYPERBOLA 


(79)  Prop.  I.     Theorem. 

The  squares  of  ordinates  to  the  transverse  axis  of  an  hyperbola  are 
to  each  other  as  the  rectangles  of  the  corresponding  abscissas. 

That  is,     GP  :  RV2    ::  AF.FB  :  AV.VB. 
Or,  GP  :  R'V'2  ::  AF.FB  :  AV'.V'B. 


(Fig.  27.) 


Through  the  vertices  A  and  B 
draw,  from  the  focal  tangent  IT, 
the  lines  LA  and  IB  at  right 
angles  to  AB ;  through  the  focus 
F  draw  LF  and  IF,  meeting  LM 
and  IK  in  M  and  K;  and  pro- 
duce RV  both  ways  to  P  and  N. 

By  the  principles  of  construc- 
tion (15^), 

IB=BF,    and  AL=AF. 


But'    IB  :  BF  ::  SV  :  FV,  and  AL  :  AF  ::  VN  :  FY. 

Therefore     SV=FV,  and  VN=FV. 
By  similar  triangles  (LGF  and  LPN)  we  have 

GF  :  PN  ::  LG  :  LP  ::  »»AF  :  AV; 
and  by  similar  triangles  (GIF  and  PIS)  we  have 

GF  :  PS  ::  GI  :  PI  ::  »'FB  :  VB. 


•  Leg.  4.  18.    Euc.  6.  4. 
»  L.,4,  19;  W.,  3,  4  and  5. 


*  Leg.  4.  15.  Cor.  2.    Euc.  6.  3, 
b  L.,4, 16;  W.,  3,  2. 


OF    THE    HYPERBOLA. 


45 


Multiplying  the  proportions  together, 

GF2  :  PN.PS  ::  AF-FB  :  AV-VB. 

But    PS=PV-SV=(by2)FR-SV=FK-FV; 

and    PN=PV  +  VN=(by  2)  FR+VN=:FK+FV. 

Therefore      PN-  PS==FR  +  FV.  FR^FV="FE2— FV2=bRV2. 

Substituting  this  value  of  PN-PS,  we  have 

GF2  :  RV2  ::  AF-FB  :  AV-VB. 

By  using  the  accented  letters,  P',  R',  S',  V,  W,  the  foregoing 
demonstration  will  apply  to  the  other  branch  of  the  hyperbola. 

(80)  Cor,  1.  Hence,  if  two  ordinates  are  equally  distant  from 
the  centre,  they  are  equal  to  one  another. 

That  is,    if  CFr=CV,  then  GF=RV. 

(81)  Cor.  2.  Hence,  the  two  portions  of  the  curve  lying  on  the 
opposite  sides  of  the  transverse  axis  AB,  or  the  conjugate  axis 
DE,  are  similar;  and  if  placed  upon  one  another,  would  coincide 
in  every  part.  For  if  at  any  point  they  should  not  coincide,  the 
ordinates  at  that  point  would  be  unequal,  which  by  Cor.  1  is  im- 
possible. 

(82)  Cor.  3.  Hence,  there  is  another  point  situated,  in  respect 
to  the  curve,  precisely  like  the  focus  F,  and  may,  therefore,  be 
called    another   focus.    Thus, 


if  CV=CF,  the  point  V  is  the 
other  focus. 

(83)  Cor.  4:.  If  different  hy- 
perbolas have  the  same  trans- 
verse axis,  the  corresponding 
ordinates  are  proportional  to 
each  other.     That  is, 

FH  :  FH'  ::  OS  :  GS'. 


(Fig.  28.) 


Leg.  4.  10.     Euc.  2.  5,  Cor. 
L.,4,  10;  W.,§377. 


b  Leg.  4. 11.    Euc.  1.  47. 
b  L.,4, 11;  W.,4,  8. 


46 


CONIC  SECTIOISft. 

For  (79) 

FH^ 

:  GS2    ::  AF.FB  :  AG.GB 

Also, 

FH2 

GS'2  ::  AF.FB  :  AG.GB. 

Therefore 

FH'-^ 

FH'^::  GS2        :  GS'^. 

And* 

FH 

FH'  ::  GS          :  GS'. 

(84)  Cor.  5.     It  follows  from  the  last  of  tlic  foregoing  propoi 
tions,"  that  HH'  :  SS'  ::  FH  :  GS. 

(85)  Cor.  6.     Since  OC  is  midway  between  AL  and  BI,  lyii.g 
on  opposite  sides  of  AB,  it  equals  half  their  difference. 

But  BI-AL=BF-AF=AB. 

Therefore  OC  is  equal  to  AC,  the  semi- transverse  axis. 

(86)  Prop.  II.     Theorem. 

The  square  of  an  ordinate  to  the  transverse  axiSy  is  to  the  rectangle 
of  the  corresponding  abscissas,  as  the  square  of  the  conjugate  axis 
is  to  the  square  of  the  transverse  axis. 

That  is,     RV2  :  AV.VB  : :  DE^  :  ABl 

(Fig.  29.) 


For  by  similar  triangles  (IGF  and  ILM)  we  have 

GF  :  LM=2AF  ::  IG  :  IL  ::  FB  :  AB. 

And  by  similar  triangles  (LGF  and  LIK)  we  have 
GF  :  IK=2FB  ::   LG  :   IL  ::  AF  :  AB. 


Leg.  2.  12,  Cor. 
L.,2,  11;  W.,  3,  10. 


••  Leff.  2.  6.     Enc.  6,  D.  and  16. 
b  L.,  2,  6  and  7 ;  W.,  3,  6  and  7. 


OF  THE  HYPERBOLA. 


47 


Multiplying  the  proportions  together, 

GF2  :  4AF.FB=(7)  DE^  ::  AF.FB      AB» 
But  (79)        GF2  :  RV2  ::  AF.FB  :  AV.VB. 
Therefore,*  RV^  :  AV.VB  ::  DE^  :  AB^. 

(87)  Cor.  1.     The  line  GH  is  the  parameter  of  the  transverse 
axis. 
For,  as  above,     GF^  :  DE^  ::  AF.FB=iDE2  :  ABl 
Therefore,  extracting  roots,'' 

AB  :  iDE  ::  DE  :  GF=iGH. 
Or,  doubling  the  second  and  fourth  terms, 
AB  :    DE   ::  DE  :  GH. 


(88)  Cor.  2.  If  a  circle  be  de- 
scribed on  the  transverse  axis  of  an 
nyperbola,  an  ordinate  to  the  hyper- 
bola is  to  a  tangent  to  the  circle 
drawn  from  the  foot  of  the  ordinate, 
as  the  conjugate  axis  is  to  the  trans- 
verse. 

For  (86) 

RV2  :  AV.VB=«VN2  : 

Therefore,    ''RV  :  VN  ;; 


(Fig.  30.) 


DE2 
DE 


ABl 
AB. 


(89)  Cor.  3.  If  the  conjugate  axis  of  an  hyperbola  is  equal  to 
the  transverse,  the  hyperbola  is  said  to  be  equilateral^  and  the 
square  of  the  ordinate  becomes  equal  to  the  rectangle  of  the  cor- 
responding  abscissas. 

(89a)   Cor.  4.     CF«-CAa=CDl 

For    '^CF2-CA2=AF.FB=(7)CD«. 


Leg.  2.  4.     Euc  6.  24. 

Leg.  4  30 

L.,  2,  4,  Cor. 

L.,  4,  29  ;  W.,  3,  13. 


"  Leg.  2.  12,  Cor. 

"  Leg.  4.  10.     Euc.  2.  6,  Cor. 

b  L.,3,  11;  W.,  3,  10. 

d  L.,4,  10;  W.,  §377. 


48 


CONIC  SECTIONS. 


(90)  Prop.  III.     Theorem. 

The  difference  of  two  lines  drawn  from  the  foci  of  an  hyperbola  to 
any  point  in  the  curve,  is  equal  to  the  transverse  axis. 


That  is,     VM-FM=AB. 


Make  the  arc  BN  equal  to  (Fig.  31.) 

AM,  join  FN,  draw  the  ordi- 
nates  NR  and  MS,  the  con- 
iugate  axis  DE,  and  the  focal 
tangent  GTP,  and  produce 
NR,  MS,  and  CD  to  G,  P, 
and  O. 

Then,  since  CO  is  midway 
between  GR  and  PS,  lying  on 
opposite  sides  of  AB,  it  is  equal 
to  half  their  difference. 

Therefore  GR-PS=20C=(85)  AB. 

But  (2)     GR-PS=FN-FM=(81)  VM-FM 

Therefore  VM-FM=AB. 

(91)  Scholium.  The  property  proved  in  this  proposition  fur- 
nishes the  definition  of  the  hyperbola  in  many  treatises.  It  also 
affords  a  ready  method  of  describing  the  curve  mechanically.  Take 
a  thread  and  a  ruler,  such  that  the  excess  of  the  length  of  the  ruler 
over  that  of  the  thread  shall  be  equal  to  the  transverse  axis,  and 
the  sum  of  their  lengths  greater  than  the  distance  between  the  foci. 
Fasten  one  end  of  each  together,  and  the  other  ends  one  to  each 
focus.  Place  a  pencil  against  the  thread,  and  press  it  against  the 
ruler  so  as  to  keep  it  constantly  stretched,  while  the  ruler  is  turned 
around  the  focus  to  which  it  is  attached  as  a  centre.  The  point 
of  the  pencil  will  describe  one  branch  of  the  hyperbola.  For  in 
every  position  of  the  pencil,  the  difference  of  the  distances  to  the 
foci  will  be  equal  to  the  difference  between  the  length  of  the  ruler 
and  that  of  the  string. 


OF  THE  HYPERBOLA. 


49 


(92)  Prop.  IV.     Theorem. 

7W  lines  drawn  from  the  foci  to  ony  poim  in  the  curve,  make  equal 
angles  witl  a  fangfrtit  to  the  curve  at  that  point. 

That  is,  FM  and  VM  make  equal  angles  with  the  tangent  TU, 
or  FMT=VMT. 


If  not,  let  them  make  equal  (Fig.  32.) 

angles  with  RS,  so  that  FMR 

=:VMR. 

Since  there  cannot  be  two 
different  tangents  to  a  curve  at 
the  same  point,  RS  must  cut  it, 
and  fall  within,  as  at  some  point 
E.  With  the  centre  F  and  ra- 
dius FE  describe  the  arc  EK, 
cutting  the  curve  in  K.  Cut 
off  MG=MV,  and  join  GV,  GE, 
EF,  EV,  KF,  and  KV. 

The  angle  RMF=(by  supposition)  VMR.  Then,  in  the  triangles 
GMD  and  VMD,  GM=My,  and  MD  is  common,  and  the  angle 
GMD=VMD ;  therefore  GD= VD,  and  the  angle  GDM=VDM. 
Hence,  in  the  triangles  EGD  and  EVD,  the  sides  GD  and  ED= 
VD  and  ED,  and  the  angle  GDE=VDE  ;  therefore  EG=EV 
But  EV  is  less  than  KV,  because  the  angle  EFV  is  less  than  KFV. 
while  the  sides  EF  and  FV  are  equal  to  KF  and  FV.*  Therefore 
EG  is  less  than  KV,  and  consequently  EF— EG  is  greater  than 
KF  -  KV.  Now  (90)  KF  -  KV=  FM  -  MV  =  FM  -  MG  =  FG. 
Hence  EF— EG  is  greater  than  FG.  Or,  adding  EG  to  both,  EF 
is  greater  than  FG+EG.  That  is,  one  side  of  a  triangle  is  greater 
than  the  sum  of  the  other  two  sides,  which  is  impossible.''  In  the 
same  manner  it  may  be  shown,  that  no  other  line  but  TU  makes 
equal  angles  with  FM  and  MV,  and  consequently  TU  does. 


•  Leg.  1.  9.    Euc.  1.  24. 

•  L.,  1,13;  W.,1,31. 


"  Leg.  1.  7.    Euc.  1.  20. 
«»  L.,  1,  19  ;  W.,  1,  26. 


>v 


»^  V 


b^ 


CONIC  SECIIONS. 


(93)   Co?'  1.     Hence,  to  draw  a  tangent  to  the  curve  at  any 
point  M,  join  MF  ana  MV,  and  bisect  the  included  angle  VMF 

(£>4)  Co**.  *2.     GV  i=>  perpendicular  to  TU   and  is  bisected  at  the 
point  L. 

(95)   Cor.  H      FG-^AB,  since  each  is  equal  to  FM— MV. 


(96)  Prop.  V.     Theorem. 

If  a  line  he  drawn  from  either  focus  perpendicular  to  a  tangent  t^t 
the  curve  at  any  point,  the  distance  of  its  intersection  from  the 
centre  is  equal  to  the  stmi-transversc  axis. 


(Fig.  33.) 


u     I 


That  is,     CL=AC. 

Since  FV  is  bisected  in  C, 
and  (by  the  preceding  propo- 
sition) GV  in  L,  CL  is  par- 
allel* to  FG,  and  the  triangles 
LCV  and  GFV  are  similar. 
Hence 

CV  :  FV  ::  CL  :  GF. 
But  CV==iFV, 


Therefore     CL=JGF=(95)  JAB=AC. 

(97)  Cor.  1.  Hence,  a  circle  described  on  the  transverse  axis 
with  the  centre  C,  will  pass  through  the  intersections  L  and  P , 
and  conversely,  if  from  any  point  in  the  circumference  of  such  a 
circle,  two  lines  be  drawn  at  right  angles  to  one  another ;  and  if 
one  of  them  pass  through  one  of  the  foci,  the  other  will  be  a  tan- 
gent to  the  curve. 


•  Leg.  4.  16.    Euc.  6.  2. 

»  L.,  4,  16,  converse  ;  W.,  3,  3. 


OF  THE  HYPERBOLA. 


51 


(98)   Cor.  2.     Hence,  a  diameter  HI,  parallel  to  TU,  would  cut 
off  a  part  MX  of  the  line  MF,  equal  to  the  semi-transverse  axis 

For*    MX=CL=AC. 


(99)  Prop.  VI.     Theorem. 

The  rectangle  of  the  two  perpendiculars  drawn  from  the  foci  to  the 
tangent  to  the  curve  at  any  point,  is  equal  to  the  square  of  the 
semi-conjugate  axis. 


(Fig.  34.) 


That  is,     VL.FP=CD2. 

Join  PC,  and  produce  PC  and 
LV  till  they  meet  in  N. 

Then  will  the  triangles  PFC  and 
NVC  be  similar  and  equal,  since 
the  angle  PCF=NCV,  and  FPC- 
the  alternate  angle  VNC,  and  the 
side  FC=VC.  Therefore  CN=CP, 
and  (97)  the  point  N  is  in  the  cir- 
cumference of  a  circle  described  on 
AB  as  a  diameter.  Consequently,** 
NV.VL=AV.VB.     Or,  since  NV=PF,  PF.VL=AV.VB=(7)  CD» 


(100)  Prop.  VII.     Theorem. 

If  at  any  point  in  the  curve  a  tangent  and  ordinate  be  drawn,  meet- 
ing either  axis  produced,  half  that  axis  is  the  mean  proportional 
between  the  distances  of  the  two  intersections  from  the  centre. 

That  is,     CS  :  CA  ::  CA  :  CT. 
Or,  CG  :  CD  ::  CD  :  CH. 


Leg.  1.  28.     Euc.  1.  34. 
L.,1,  30;  W.,  1,  39. 


"  Leg.  4.  29,  Cor.     Euc.  3. 
»>  L.,4,38,  Cor.;  W.,  3.  11. 


52 


CONIC  SECTIONS. 


Connect  M  with  the  foci  (Fig.  36.) 

F  and  V,  draw  VL  perpen- 
dicular to  the  tangent,  and 
join  CL  and  SL. 

Since  MSV  and  MLV 
are  both  right  angles,  each 
is  an  angle  in  a  semicircle,* 
and  consequently,  a  circle 
described  on  MV  as  a  di- 
ameter, would  pass  through 

L  and  S.     Then  must  the  angles  VML  and  VSL  be  equal,  being 
in  the  same  segment,^  or  measured  by  the  same  arc. 

But  (92)  VML  =  FML  =  CLT,  since  CL  is  parallel  to  FM. 
Therefore  the  angle  VSL  or  CSL=CLT.  Hence,  the  triangles 
LCT  and  SCL  are  similar,  for  the  angle  CSL=CLT,  and  the  angle 
at  C  is  common. 

Therefore     CS  :  CL  ::  CL  :  CT. 
But  (96)  CL=CA. 

Therefore     CS  :  CA  ::  CA  :  CT, 
which  proves  the  proposition  in  respect  to  the  transverse  axis. 

Again,  since  when  three  numbers  are  in  continued  proportion, 
the  first  is  to  the  third  as  the  square  of  either  antecedent  is  to  the 
square  of  its  consequent,  we  have  from  the  last  proportion 

CS2  :  CA2  ::  CS=GM  :   CT  ::  (sim.  tri.)  GH  :  CH. 
Hence,  by  division,^    CS^-CA^  :  CA^  ::  CG  :  CH. 
But  (86)       AS.SB='^CS2-CA2  :  CA'  ::  MS^^CG^  :  CD«. 
Therefore,  by  equality  of  ratios, 

CG  :  CH    ::  CG^  :  CD^ 
which  gives,     CG.CH=CDl 

Or.     CG  :  CD    ::  CD    :  CH. 


•  Leg.  3.  18,  Cor.  2.    Euc.  3.  31. 

•  Jjeg,  2.  6.    Euc.  6.  17. 

•  L.,3, 15,  Cor.;  W.,  2,  14,  Cor. 
«^  L.,  2,  6  and  7;  W.,  3,  6  and  7. 


"  Leg.  3.  18.     Euc.  3.  21. 

"  Leg.  4.  10.    Euc.  3.  6,  Cor. 
>  L.,  3,15;  W.,  2,  14. 
d  L.,4,  10;  W.,8  377. 


OF  THE  HYPERBOLA. 


58 


(101)  Prop.  VIII.     Theorem. 

[f  different  hyperbolas  have  the  same  transverse  axis,  the  corre^ 
sponding  sub-tangents  are  equal  to  one  another. 

Let   EAD,    E'AD',    &c.    be  (Fig.  36.) 

any  number  of  hyperbolas, 
each  having  AB  for  its  trans- 
verse axis,  SG  produced  any 
ordinate  to  each,  and  ST, 
S'T,  &c.  tangents  at  the  points 
v^here  the  ordinate  meets  the 
curves. 

Then  for  each  of  them  we 
have  (100) 

CG  :  CA  ::  CA  :  CT, 

and  since  the  first  three  terms  are  the  same  for  all,  the  fourth  must 
be  likewise,  which,  subtracted  from  CG,  gives  the  sub-tangent. 


(102)  Cor.  1  Hence,  we  may  draw  a  tangent  at  a  given  point 
in  the  curve,  without  knowing  the  foci. 

Let  S  be  the  given  point.  On  AB  describe  a  circle ;  draw  the 
ordinate  SG  and  the  tangent  to  the  circle  GN,  let  fall  the  perpen- 
dicular NT,  and  join  TS. 

For  in  the  right-angled  triangle  CNG,  we  have 

CG  :  CN  ::  CN  :  CT. 

But  CN=CA;   therefore  CG   :   CA  ::  CA  :  CT,  which  is  the 

same  proportion  as  the  one  above,  showing  that  the  ordinate  at  N 
and  the  tangent  at  S  meet  the  transverse  axis  in  the  same  point. 

(103)  Cor.  2.     CG.GT=AG.GB,  each  being  equal*  to  NG«. 


'  Leg.  4.  23  and  30.     Euc.  6.  8,  Cor.  and  3.  36. 
»  L.,  4,  23  ;  W.,  2,  14. 


54 


CONIC  SECTIONS. 


(104)  Prop.  IX.     Theorem. 

The  difference  of  the  squares  of  two  ordinates  drawn  to  either  axis 
from  the  extremities  of  any  two  conjugate  diameters^  is  equal  to 
the  square  of  half  the  other  axis. 

That  is,     FU3-GF=AC2,    and    PS^- UR«=CE". 
(Fig.  37.) 


Draw  the  tangents  PT  and  UK,  meeting  the  transverse  axis  in 
T  and  V,  and  the  conjugate  axis  in  H  and  K.  Then  CS.CT= 
CR.CV,  each  being  equal  (100)  to  AC^  or  BCl 

Therefore'^     CS  :  CR  ::  CV  :  CT. 

But,  since  (9)  UV  is  parallel  to  PC,  and  UC  to  PT,  the  tri 
angles  PTC  and  UCV  are  similar;  as  also  PCS  and  UVR. 
Hence     VR  :  CS  ::  UV  :  PC  ::  CV  :  CT. 
Then,  by  equality  of  ratios,  we  have 

CS  :  CR  ::  VR  :  CS, 
And,  consequently,     CR.VR=CSl 
But  (103)      CR.VR=AR.BR=''CR2-.AC2. 
Therefore      CR>~AC2=CS2;    or,  CR2-CS«=AC«. 
That  is,         Fm-GP?=ACl 


•  Leg,  2.  2.     Euc.  6.  16. 

*  L..2.  2:  W..  3.3. 


"  Leor.  4.  10.    Kuc.  2.  6,  Cor. 
b  L.,  4,  10;  W.,  §3?7. 


OF  THE  HYPERBOLA. 


55 


By  comparing  the  similar  triangles  UKC  and  PCH,  and  also 
UCF  and  PHG,  it  may  be  proved  in  tlie  same  manner  that  PS'^— 

{104:0)    AK.BK=CS^  each  being  equal  to  CR.YR. 

(105)  Prop.  X.     Theokem. 

TTie  difference  of  the  squares  of  any  pair  of  conjugate  diameters  o, 
an  hyperbola  is  equal  to  the  difference  of  the  squares  of  the  two 
axes. 

That  is,    UN2-PL3=AB«-DE«. 
(Fig.  87a.) 


Foi*  UC«~PC«=CR«-CS«+UR«-.PS«=CR«-CS«-(PS«-.UR«) 
But  (104)    CR«-CS2=AC«,  and  PS*-UR«=CE«. 
Therefore    UC2-PC2=AC«-CE2. 
Or,  UN»-PL2=AB2-DE». 


'  Leg.  4.  11.    Euc.  1.  47. 

»  L,  4,  11;  W.,  4,  8. 


56 


CX)NIC  SECTJONS, 


(106)  Prop.  XI.     Theorem. 

Any  parallelogram  inscribed  between  conjugate  hyperbolas,  hamng 
its  sides  parallel  to  two  conjugate  diameters,  is  equal  to  the  rect- 
angle of  the  two  axes. 

That  is,    GHIK=AB.DE. 

(Fig.  38.) 


Draw  CX  at  right  angles  to  TK ;  then  will  the  triangles  TCX 
and  CUR  be  similar. 

Now    DC«  :  AC2  ::  UR^  :  AR.RB={104^)  CS^. 

On  extracting  roots,  DC  :  AC  ::  UR  :  CS. 

Or,  alternately,  CS   :  AC  ::  UR  :  DC. 

But  (100)  CS   :  AC  ::  AC  :  CT. 

Therefore  UR  :  DC  ::  AC  :  CT, 

or  UR.CT=AC.DC=iAB.DE. 

Also  (sim.  tri.)  UR  :  CU  ::  CX  :  CT, 

or  UR.CT=CU.CX'^=iGHIK. 

Therefore  GHIK=AB.DE. 

(107)  Cor,    AC  :  CX  ::  CU  :  DC,    or  CU.CX=AC.DC. 

•  Leg.  4.  5. 

«'L.,4,5;  W.,4,5. 


OF  THE  HYPERBOLA. 


m 


(108)  Pkop.  Xn.     Theobem. 

The  rectangle  of  ivjo  lines^  d/rawn  from  the  fod  to  a/ivy  jpomt  in 
the  Gurve^  is  equal  to  the,  square  of  half  the  diameter  jparallel  to 
the  tangent  at  thatjpoint. 


That  is,     FMxVM-CHa. 


Draw  the  tangent  MLP,  and  (Fig.  39.) 

the  perpendiculars  to  it  FL, 
MN,  and  VP.  Then  the  tri- 
angles PMV,  SMN,  and  FML 
will  be  similar  (92). 

Therefore 
MS  :  MN  ::  FM  :  FL. 

And  also, 
MS  :  MN  ::  VM  :  VP. 

Multiplying  the  proportions  together, 

MS2  :  M.W  ::  FMxVM  :  FLxVP. 
But  (98)  MS2=AC2,     and  (99)  FLxVP=CB«. 

Therefore  AC^  :  MN^  ::  FMxVM  :  CE«. 

Now  (107)  MN.CH=AC.CE. 

Therefore         AC    :  MN    ::  CH    :  CE. 
Or,  squaring,     AC^  :  MN^  ::  CH^  :  CEl 
Hence,  by  equality  of  ratios,    CH^  :  CE' 
And     FMxVM=CH". 


FMxVM  :  CE^ 


(109)  Prop.  XIII.     Theorem. 

If  at  one  of  the  vertices  of  an  hyperbola  a  tangent  he  dravm  meet 
ing  any  diameter ^  and  also  from  the  same  point  an  ordinate  to 
thai  diameter  produced,  the  semi-diameter  is  the  mean  propor- 
tional  between  the  distances  (f  the  two  intersections  from  the 
centre, 

8 


58 


CONIC  SECTION& 

That  is,    CE  :  CI  ::  CI  ;  CP. 

(Fig.  40.) 


Draw  IT  and  10  tangent  and  ordinate  to  the  curve  at  I. 

Then  (sim.  tri.)  CE  :  CI  ::  CB  :  CT  ::  (100)  CO  :  CB  ::  (sim 
tri.)  CI  :  CP. 

(110)  Cor.  1.  Hence  CP  :  CP^  ::  CE  :  CP.  Also,  if  the  ordi. 
nate  BS  be  drawn  parallel  to  HI,  we  have  CN^  :  CR^  ::  CS  :  CR. 

For  (sim.  tri.)  CN^  :  CR^  ::  CF2=(as  shown  in  104)  CO.OT  . 
CB2=(100)CO.CT  ::  OT  :  CT  ::  (lll)EP  :  CP  ::  (sim.  tri.)  BE 
=CS  :  CR. 

(111)  Cor.  2.  The  lines  CE  and  CO  are  similarly  divided,  the 
former  in  I  and  P,  and  the  latter  in  B  and  T ;  and,  consequently, 
lines  joining  EO,  BI,  and  PT,  would  be  parallel. 

(112)  Cor.  3.  The  triangle  CIT=CBP,  and  CEB=C01,  for 
the  angle  at  C  is  common,  and  the  sides  about  it  reciprocally  pro- 
portional.* 

(113)  Cor.  4.     The  area  of  triangle  CNG=CBP  or  CIT. 

For''  CNG  :  CRP  ::  CN*  :  CR^  ::  (Cor.  1)  CS=BE  :  CR  :: 
(sim.  tri.)  BP  :  PR  ::  ^'CBP  :  CRP. 

Hence,  since  CNG  and  CBP  have  the  same  ratio  to  CRP,  they 
are  equal  to  one  another. 


Leg.  4.  24,  Cor.     Euc.  6.  16. 
Leg.  4.  6,  Cor.    Euc.  6.  1. 
L,  4,  24;  W.,  4,  13. 
L.,4,  6,  Cor.  1;  W.,  4  5,  Cor.  2. 


"  Leg.  4.  26.    Euc.  6.  19. 
b  L.,4,  25;  W.,  4, 14. 


OF  THE  HYPEilBOLA.  59 

<114)   Cor.  5.     IOBP=IOT. 

For.  (112)  CIT=CBP,  and  taking  each  from  CIO,  the  remainders 
must  be  equal 

(115)   Cor.  6.     The  triangle  UZX=PBXL,  (Z  being  any  point 
in  the  curve.) 

For      CB  :  BP  ::  CX  :  XL  ::  '<:;B+CX=AX  :  BP+XL. 

Also,    CB  :  BP  ::  CO  :  01     ::    CBH-CO=AO   :  BP+OI. 

Thereiore  AX  :  AO  ::    BP+XL  :  BP+OL 


Or,''      AX.XB  :  AO.OB  ::  BP+XL.XB  :  BP-fOLOB. 
But 


BP+XL.XB=''2PLXB,     and    BP+OI.OB=2PIOB=(114)  210T 

Therefore 
PLXB  :  lOT  ::  AX.XB  :  AO.OB  ::  (79)  ZX*  :  OP  ::  <»UZX  :  lOT 

Hence,     PLXB=UZX,  since  both  have  the  same  ratio  to  lOT. 

(116)   Cor.  7.     DZL=DUC-ICT. 

For  (112)  ICT  =  CBP  =  CLX-PLXB  =  (115)CLX-.UZX  = 
DUC— DZL,  since  the  part  LDUX  is  common  to  both  CLX  and 

uzx. 

Hence,     DZL=DUC-ICT. 


(117)   Prop.  XIV.     Theorem. 

The  square  of  any  diameter  is  to  the  square  of  its  conjugate,  as  the 
rectangle  of  its  abscissas  is  to  the  square  of  the  corresponding 
ordinate. 

That  is,     HP  :  MN^  ::  HD.DI  :  DZl 


•  Leg.  2.  10.     Euc.  6.  12.  "  Leg.  2.  8.     Euc.  6.  16  and  16. 
'  Leg.  4.  7.  "  Leg.  4.  25.     Euc.  6.  19. 

•  L.,2,  £;  W.,  3,8.  "  L.,  2, 10;  W,  3, 11. 
«  L.,  4,  7;  W.,  4,  6.  ^  L.,  4,  25 ;  W.,  4,  14. 


60 


CONIC  SECTIONS. 
(Fig.  41.) 


For*  CNa  :  DZ^  ::  CNG=(113)ICT  :  DZL=(116)DUC-1CT 

But*  ICT  :  DUG  ::  CP  :  CD^. 

Hence,''     ICT  :  DUC-ICT  ::  CP  :  CD2-CP=«HD.DI. 

Therefore,  by  equality  of  ratios, 


CN3 

:  DZ2    : 

:  CP  :  HD.DI. 

Or,  alternately,     CP 

:  CN2   : 

:  HD.DI  :  DZl 

Or,-^                       HP 

:  MN^  : 

:  HD.DI  :  DZl 

(118)  Cor,  Hence,  all  chords  parallel  to  any  diameter  are  bi- 
si»eted  by  its  conjugate ;  and,  conversely,  a  line  bisecting  two  or 
more  parallel  chords  is  a  diameter. 


(119)  Prop.  XV.     Problem. 

To  draw  a  tangent  to  an  hyperbola  from  a  given  point  without 

the  curve. 

Let  AMBM'  be  the  given  hyperbola,  AB  the  transverse  axis,  F 
one  of  the  foci,  and  T  the  given  point. 

Join  TF,  and  upon  it  and  AB  as  diameters,  describe  the  circles 
TPF  and  APB,  cutting  each  other  in  P  and  F.  The  lines  TPM 
and  P'TM'  drawn  through  the  points  of  intersection,  will  be  tan- 
gents  to  the  hyperbola. 


•  Leg.  4.  25.     Euc.  6.  19. 

•  Leg.  4.  10.     Euc.  2.  5,  Cor. 
»  L.,  4,  25;  W.,4, 14. 

<=  L..  4,  1    ;  W,  §  377. 


*Leg.  2.  6.    Euc.  6,D. 

""  Leg.  2.  7.    Euc.  6.  16. 

b  L,  2,  6  and  7;  W  ,  3,  6  and  7. 

d  L.,2,  10;  W,  3,  11. 


OF  THE  HYPERBOLA. 


61 


Join  FP.  The  angle  FPT 
is'  a  right  angle,  and  FP  per- 
pendicular to  TM.  Now,  since 
from  the  point  P,  in  the  cir- 
cumference of  the  circle  de- 
scribed on  the  transverse  axis, 
there  are  drawn  two  lines,  PF 
and  PM,  at  right  angles  to  one 
another,  and  one  of  them  (PF) 
passes  through  the  focus,  the 
other  must  (97)  be  tangent  to  the  hyperbola. 


(Fig.  42.) 


(120)  Prop.  XVI.     Problem. 
To  find  the  centre,  axes,  and  foci  of  a  given  hyperbola. 


rFicr.  43.) 


Let  AKBL  be  the  given  hyperbola. 

Draw  any  two  pairs  of  paraillel  chords  HI  and  LK,  MN  and  OP, 
bisect  them  in  T,  U,  R,  and  S ;  join  TU  and  RS,  and  produce 
the  lines  till  they  meet  in  C.  Both  these  lines  being  (118)  diamet^iis, 
the  point  C  must  be  the  centre. 

From  the  centre  C,  and  with  any  convenient  radius,  describe  a 


•  Leg.  3.  18,  Cor.  2.     Euc.  3.  31. 

•  L.,  3,  15,  Cor.  8 ;  W., 2,  14,  Cor. 


62 


CONIC  SECTIONS 


circle  cutting  the  hyperbola  in  any  points  W  and  X.  Join  WX^ 
and  at  right  angles  to  it  draw,  through  the  point  C,  the  line  ABY. 
Since  this  line  bisfects  *WX  at  right  angles,  it  divides  the  curve 
into  two  similar  parts,  and  therefore  (81)  AB  is  the  transverse 
axis. 

To  find  the  foci,  draw  a  tangent  at  any  point  S'  in  the  hyperbola, 
and  from  the  point  T^  where  it  intersects  a  circle  described  on  the 
transverse  axis,  draw  T'V  perpendicular  to  it.  The  point  V  is 
(97)  one  of  the  foci.  From  V  draw  VZ  tangent  to  the  circle 
AT'B,  and  from  C  draw  CD  and  CE  at  right  angles  to  AB,  making 
each  equal  to  VZ.  The  tangent  VZ  is  ''the  mean  proportional  be- 
tween AV  and  VB,  therefore  (7)  DE  is  the  conjugate  axis. 


(120a)   Prop.  XVII.     Theorem. 

An  hyperbola  may  he  formed  by  the  mutual  intersection  of  a  cone 

and  a  plane. 

Let  CDLE  and  CM'U'N'  represent  the  two  nappes  of  a  cone, 
and  CDE  and  CM'N'  tw^o  triangular  sections  formed  by  a  plane 
perpendicular  to  the  base,  and  passing 
through  the  vertex  of  the  cone.  Let  the 
hyperbola  in  Prop.  I.,  with  no  change  of 
letters,  be  placed  one  branch  upon  each 
nappe,  in  the  manner  of  a  collar,  with  its 
plane  perpendicular  to  that  of  the  triangu- 
lar sections,  and  the  vertices  A  and  B  in 
contact  with  the  surface  of  the  cone  some- 
where on  the  lines  CM'  and  CD.  Further, 
let  the  base  of  the  cone  be  so  broad  that  if 
AB  were  placed  perpendicular  to  it,  the 
hyperbolas  would  fall  within  the  cone. 

Islow,  suppose  the  point  A  to  slide  up 
or  down  on  the  line  CM',  and  B  on  CD, 


'  Leg.  3.  6.    Euc.  3.  3. 
•  L.,  3,  6  ;  W.,  2.  7. 


'  Leg.  4.  30.     Euc.  3.  36. 
^  L.,4,29;  W.,  3,13. 


OF  THE  HYPERBOLA  68 

till  the  point  G  shall  He  in  the  surface  of  the  cone;  a  condition 
which  we  shall  see  to  be  possible,  if  we  consider  that  when  A  or  B 
coincide  with  C,  the  hyperbolas  must  fall  wholly  without  the  cone. 
We  assert  that  any  other  point,  R  or  R',  in  the  hyperbola  will  also 
lie  in  the  surface  of  the  cone. 

If  net,  it  must  lie  either  within  or  without  the  cone.  Let  it  be 
supposed  to  lie  without,  and  that  the  ordinate  RV  cuts  the  surface 
of  the  cone  at  s.  Through  G  and  s  let  the  circular  sections  PGSH 
and  M5NU  be  made  to  pass,  parallel  to  the  base,  and  cutting  the 
triangular  sections  in  PS  and  MN.  The  lines  GF  and  RV  being 
perpendicular  to  the  plane  PSCDE,  must  also  be  perpendicular  to 
PS  and  MN. 

By  sim.  tri.  AFP  and  AVM,     PF  :  MV  ::  AF  :  AV. 

And  by  sim.  tri.  BFS  and  BVN,     FS  :  VN   ::  FB  :  VB. 
Multiplying  the  proportions  together, 

PF.FS  :  MV.VN  ::  AF.FB  :  AV.VB. 
But*  PF.FS=GF2,     and  MV.VN=5Va. 

Therefore        GF^  :  s\^    ::  AF.FB  :  AV.VB. 
But  (79)  GP  :  RV^  ::  AF.FB  :  AV.VB. 

Therefore        5V^=RV^,     and  5V=RV,  which  is  impossible. 

In  the  same  manner  it  may  be  shown,  that  the  point  R  cannot  lie 
within  the  cone,  and,  consequently,  it  lies  in  the  surface.  And 
since  R  is  any  point  in  the  hyperbola,  the  whole  curve  must  lie  in 
the  surface  of  the  cone. 

By  using  the  accented  letters  M',  R',  s'y  Y',  and  N ,  the  foregoing 
demonstration  will  apply  to  the  other  branch  of  the  hyperbola. 


•  Leg.  4.  23,  Cor.     Euc.  6.  13. 
»  L.,  4,  23,  Cor. ;  W.,  2,  14,  Cor.  1. 


64  CONIC  SECTIONS. 


CHAPTER  V. 

OF  THE  CURVATURE  OF  THE  CONIC  SECTIONS 


(121)  Before  entering  upon  the  subject  discussed  in  this  chaptei 
it  is  necessary  to  acquaint  the  student  with  the  doctrine  of  ultimate 
or  limiting  ratios ;  a  method  of  investigation  much  used  for  deter- 
mining the  ratio  between  quantities  that  are  not  commensurable 
with  each  other.  For  example,  if  we  wish  to  compare  the  area  of 
a  square  with  that  of  its  inscribed  circle,  we  may  first  compare  it 
with  the  area  of  a  regular  polygon  inscribed  in  the  circle ;  and  since 
the  greater  the  number  of  sides  of  this  polygon  the  nearer  will 
its  area  approach  to  equality  with  that  of  the  circle,  it  is  assumed 
that  by  increasing  them  indefinitely  the  two  areas  will  ultimately 
become  equal,  or  that  each  will  bear  the  same  ratio  to  the  area 
of  the  square.*  This  operation  involves  the  following  principle, 
the  truth  of  which  will  be  assumed  in  the  discussions  of  this  chap- 
ter, viz. : 

The  ratio  between  two  quantities  is  not  appreciably^  affected  by  add- 
ing tOf  or  subtracting  from  either,  an  indefinitely  small  part  oj 
itself. 

If  a  grain  of  sand  were  annihilated,  it  would  hardly  affect  the 
ratio  which  the  weight  of  the  whole  earth  bears  to  that  of  the 
moon,  or  any  other  body ;  but  even  this  would  be  far  greater  than 
'n  the  cases  in  which  we  employ  limiting  ratios. 

•  Leg.  6.  8.    Euc.  Sup.  1.  4. 

»  L.,  6,  7  ;  W.,  5,  9. 

^  Strictly  it  is  not  affected  at  all ;  for  in  the  limit  these  indefinitely  small 
quantities  are  really  nothing.  We  employ  the  language  for  the  sake  of  con- 
venience. 


OF  CLiRVATURE. 


65 


(122)  Def.  The  curvature  of  a  curve  is  its  deviation  from  a 
.angent  line,  measured  by  the  subtense  of  an  indefinitely  small  arc. 
Thus  the  curvatures  of  the  two 
curves  AG  and  AE  at  the  point 
A,  are  to  each  other  as  the  sub- 
.enses  BD  and  BC  of  the  in- 
definitely small  arcs  AD  and 
AC. 


(123)  Def.  If  in  any  curve  three  points  le  taken  at  equal  dis- 
tances, but  indefinitely  near  each  other,  the  circle  v^hich  passes 
through  them  *is  called  an  osculating  circle,  and  through  the  indef 
initely  small  arcs  lying  between  those  points,  the  two  curves  may 
be  considered  to  coincide,  that  is,  to  touch  one  another.  And  fur- 
ther, since  curves  may  be  regarded  as  polygons  of  an  indefinite 
number  of  sides,  the  parts  of  the  curves  lying  between  contiguous 
points  thus  taken  may  be  considered  straight  lines. 

(124)  Def.  The  radius  of  the  osculating  circle  is  called  the 
radius  of  curvature  of  the  curve  at  the  point  of  contact,  and  its 
diameter  the  diameter  of  curvature.  Also  any  chord  that  passes 
through  the  point  of  contact  is  called  a  chord  of  curvature.  The 
curvature  of  a  curve  may  be  determined  by  the  radius  of  curvature. 


(125)  Prop.  I.     Theorem. 

The  radius  of  curvature  at  the  vertex  of  a  conic  section,  is  equa, 
half  the  principal  parameter. 

Let  GAH  be  the  conic  section,  AB  its  transverse  axis,  OMN  the 
osculating  circle  at  the  vertex  A,  and  AM  an  indefinitely  small  arc 
common  to  both  curves. 

Put  the  parameter  equal  to  P,  and  draw  the  ordinate  MS. 


•  Leg.  3.  7.     Eiic.  4.  6. 

•  L.,  3,  7  ;  W.,  2,  38. 

9 


66  CONIC  SECTIONS. 

(Fig.  46— Ellipse.)  (Pig.  46— Parabola.) 


In  the  ellipse  and  hyperbola  (12), 


Hence  (SSb), 
But  (24  and  86), 
Therefore, 
Or,  alternately. 


AB    •  DE 
AB2  :  DE2 
AB2  :  DE^ 
AB    :  P 
AB    :  SB 


DE  :  P,  the  parameter. 
AB  :  P. 

ASaSB  :  SM2=«^AS.SO. 
AS.SB  :  AS.SO  ::  SB  :  SO. 
P  :  SO. 


Now,  the  nearer  the  point  M  is  to  A,  the  nearer  do  the  lines 
SB  and  SO  approach  to  equality  with  AB  and  AO,  and  in  the  limit 
at  A  the  ratio  between  them  becomes  that  of  equality  (121).  The 
last  proportion  will  then  read  AB  :  AB  : :  P  :  AO.  Consequently, 
P=AO,  or  ^P=JAO,  which  proves  the  truth  of  our  proposition  in 
regard  to  the  ellipse  and  hyperbola. 

Again,  in  the  parabola,  we  have  (60)  AS.P=SM2=  AS.SO. 
Therefore  P=:SO=(in  the  limit  at  A)  AO,  or  iP=}AO. 


'  Leg.  4.  23,  Cor.     Euc.  6.  13. 

•  L.,  4,  23  ;  W.,  2,  14. 

\ 


OF  CURVATURE. 


07 


(126)  Prop.  II.     Theorem. 

In  the  ellipse  and  hyperbola  the  chord  of  curvature  that  parses 
through  the  centre  is  equal  to  the  parameter  of  the  diameter  that 
passes  through  the  point  of  contact. 

That  is,     ML  :  HI  ::  HI  :  MO. 

(Fig.  47— EllipseO 


(Fig.  47— Hyperbola.) 


68  CONIC  SECTIONS. 

Let  N,  M,  and  G  be  the  three  points  through  which  the  oscula 
ting  circle  OMG  is  drawn  (123).     Join  MC  and  produce  it  to  O; 
join  OG,  produce  NM  to  T,  and  from  G  draw  GS  parallel  to  MT. 

The  triangles  MGS  and  MGO  are  similar,  for  the  angle  at  M  is 
common,  and  SGM=GMT='^MOG.     Therefore 

MO  :  MG  ::  MG  :  MS. 

And,  since  by  the  definition  of  an  osculating  circle  (123),  MG  is 
but  an  indefinitely  small  part  of  MO,  MS  must  be  but  an  indef- 
initely small  part  of  MG,  and  mucli  more  then  must  MS  be  but  an 
indefinitely  small  part  of  MO.  Consequently  (121)  the  ratio  of 
MO  to  SO,  and  also  that  of  ML  to  SL,  is  to  be  regarded  as  the 
ratio  of  equality.  Moreover,  since  the  arc  MG  is  indefinitely  small, 
the  chords  MO  and  OG  are  to  be  regarded  as  equally  distant  from 
Y  the  centre  of  the  circle,  and  therefore  ''equal  to  one  another. 
But  MO  :  OG  ::  MG  :  SG;  therefore  the  ratio  of  MG  to  SG, 
or  of  MG^  to  SG^,  is  that  of  equality. 

By  (54)  and  (117),  ML^  :  HP  ::  MS.SL  :  SG^  and,  dividing 
the  first  and  third  terms  by  the  equals  ML  and  SL,  and  multiplying 
them  by  MO,  we  have  ML.MO  :  HP  ::  MS.MO=(by  the  first  of 
our  proportions  above)  MG^  :  SGI  But  the  ratio  of  MG^  to  SG' 
is  that  of  equality,  therefore  ML.MO=HP;  or, 

ML  •  HI  ::  HI  :  MO. 
(127)   Cjr.    ML.MO=HP,  or  MC.MO=2CH3. 


(128)  Prop.  III.     Theorem. 

In  the  parabola  the  chord  of  curvature  that  passes  through  the  focu:i 
IS  equal  to  the  parameter  of  the  diameter  that  passes  through  the 
point  of  contact. 


'  liegr.  3.  21.     Euc.  3.  32.  *  Leg.  3.  8.    Euc.  3.  14. 

•  L.,  3,  16 ;  W.,  2, 16.  »»  L.,  3,  8  ;  W..  2,  8. 


OF  CURVATURE. 

That  is,     MS  :  SG  ::  SG  :  MR. 
(Fig.  48.) 


69 


Let  N,  M,  and  G  be  the  three  points  through  which  the  oscula- 
ting circle  OMG  is  drawn  (123).  Draw  MO  parallel  to  the  trans- 
verse axis  AB,  join  OG,  produce  NM  to  T,  and  draw  GS  parallel 
toMT. 

It  may  be  proved,  as  in  (126),  that  MS  :  MG  : :  MG  :  MO,  and 
also  that  in  the  Hmit  MG=SG;  therefore  MS  :  SG  ::  SG  :  MO. 
But  since  (61)  MO  and  MR  make  equal  angles  with  the  tangent 
TN,  they  cut  off  equal  arcs  of  the  circle,*  and  are  therefore  them- 
selves equal. 

Therefore     MS  :  SG  ::  SG  :  MR. 
(129)  Cor.    MR=4FM.    For  (76J)  MS  :  SG  ::  SG  :  4FM. 


*  Leg.  3.  21.     Euc.  3.  32  or  34. 

•  L.,  3,  16;  W.,  2,  16. 


70 


CONIC  SECTIONS. 


(130)  Prop.  IV.     Theorem. 

In  the  ellipse  and  hyperbola  the  chord  of  curvature  that  passes 
through  the  focus  is  a  third  proportional  to  the  transverse  axis, 
and  a  diameter  conjugate  to  that  which  passes  through  the  point 
of  contact. 

That  is,     AB  .  HI  ::  HI  :  MR,  drawn  through  the  focus  P 


(Fig.  49— ElUpse.) 


(Fig.  49— Hyperbola.) 


OF  CURVATURE. 


71 


Draw  MP  through  the  centre  Y  of  the  osculatjng  circle,  and 
consequently  at  right  angles  to  TM  and  HI ;  and  join  OP  and  RP. 
The  angles  MRP  and  MOP  are  "right  angles,  and  consequently  the 
triangles  MCU  and  MOP  are  similar,  as  also  MKU  and  MRP. 

Hence,     OM  :  MP  ::  MIT  :  MC,  and  OM.MC=MP.MU. 
Also,        MR  :  MP  ::  MU  :  MK=(36  and  98)  AC, 
and  MR.AC=MP.MU. 
Consequently,     MR.AC;=OM.MC, 


and  OM  :  MR  : :  AC 

MC  : 

:  AB 

:  ML. 

But  (126)     OM 

HI    : 

:  HI 

:  ML. 

Therefore     AB 

:  HI    : 

:  HI 

:  MR. 

(131)   Cor,  1.     MR-™'. 

asia)  Cor.  2.    We  have  above  MP.MU=OM.MC=(127)  2CH* 
Hence,    MY.MU=CHl 


(132)  Prop.  V.     Theorem. 

in  the  ellipse  and  hyperbola  the  squares  of  the  radii  of  curvature 
at  different  points  of  the  curve,  are  to  each  other  as  the  cubes 
of  the  rectangles  of  the  distances  of  each  from  the  two  foci. 

That  is  (Fig.  49),  putting  R  and  r  for  the  radii  of  curvature  at 

M  and  Z  .     . 

R2  :  r^  ::  FM.MV^  :  FZ.ZV*. 

For  (44  and  107),    CH.MU=AC.DC. 
And  (131a),  MY.MU=CH2. 

Therefore      AC.DC   :   CH^  ::  CH  :  MY=R. 
CW  ,  „„        CH« 


Hence     R= 


and  R2=- 


AC.DC       (AC.CD)** 

But  (45  and  108),       FM.MV=CH«. 


And,  consequently,     FM.MV  =CH«. 


•  Leg.  3.  18,  Cor.  2.     Euc.  3.  31. 

•  L.,  3,  15,  Cor. ;  W.,  3,  14,  Cor. 


72  CONIC  SECTIONS. 


Therefore    R»=    ^^V" 


(AC.DC)s 


■p7  yv 

In  the  same  manner  it  may  be  shown  that  r'=  . .  p'    ^.^ 

Therefore     R^  :  r^  ::  FMJIV^  :  FZTZV^. 

(132a)   Cor.     The  radius  of  curvature  varies  as  the  cube  of  the 
liameter  conjugate  to  that  which  passes  through  the  point  of  con- 

nXJ3 

tact.     For,  in  the  equation  R=  .  p  j^^,  tje  denominator    of  the 
fraction  is  constant ;  and  therefore  R  varies  as  CH^ 


(133)  Prop.  VI.     Theorem. 

In  the  parabola  the  squares  of  the  radii  of  curvature  at  different 
points  of  the  curve,  are  to  each  other  as  the  cubes  of  the  distances 
from  the  focus. 

T'lat  is,  putting  R  and  r  for  the  radii  of  curvature  at  M  and  Z, 
R2  :  r2  ::  FM*     FZ«. 

(Fig.  61.) 


OF  CURVATURE.  78 

Draw  MP  the  diameter  of  the  osculating  circle,  join  RP,  and 
draw  FL  perpendicular  to  MT.  Then  will  the  triangles  MRP  and 
FML  be  similar.' 

Hence  MP    :  MR=(129)  4FM  ::  FM  :  FL. 


And  squaring,     MP^  :  4FM    ::  FM^  :  FL2=;(68c)  AF.FM. 
Therefore  MP  :  iFM"  ::  FM    :  AF,     and'MP^i^^. 

In  the  same  manner  it  may  be  shown  that  the  square  of  the 
diameter  of  curvature  at  Z=         .     Therefore  the  square  of  the 

diameter  of  curvature  at  M   :   the  square  of  the  diameter  of  cur- 
vature at  Z  : :  FM'  :  FZ^  and  the  radii  will  have  the  same  ratio ; 

That  is,     R2  :  r2  ::  FM3  :  FZ«. 


(134)  Prop.  VII.     Theorem. 

If  straight  lines  he  drawn  from  one  of  the  foci  of  a  conic  section 
to  the  curve,  so  as  to  cut  off  indefinitely  small  hut  equal  sectors, 
the  curvatures  of  the  included  arcs  towards  that  focus  are  to  each 
other  inversely  as  the  square  of  their  distances  from  it. 

That  is,  Mx  :  Aw  ::  AP  :  FM^,  provided  the  areas  FAO  and 
FMG  are  indefinitely  small  but  equal. 

First,  in  the  case  of  the  ellipse  and  hyperbola.^  Let  AB  and 
DE  be  the  axes,  F  and  V  the  foci,  M  any  point  in  the  curve,  and 
KAO  and  MGRP  osculating  circles  at  the  points  A  and  M. 

It  may  be  shown,  in  the  same  manner  as  in  the  first  proportion  in 
(126),  that  Mx  :  MG  ::  MG  :  MR,  and  consequently  that  Mx.MR 
=MG2 ;  and  also  that  At^.AK= AOl 

•  Leg.  3.  21  and  18,  Cor.  2.     Euc.  3.  31  and  32. 
»  L.,  3,  16;  W.,  2,  16.     And  L.,  3,  15,  Cor.;  W.,  2,  14,  Cor. 
*>  It  is  not  thought  necessary  to  add  the  figure  for  the  hyperbola,  as  it  is  per- 
fectly analogous  to  that  for  the  ellipse. 

10 


u 


CONIC  SECTIONS. 
(Fig.  62— Ellipse.) 

N 


Therefore     Ma;.MR  :  Am^.AK 


4FM.MV 


But  (131),    MR=^=^^=(45  and  108) 

DE2  4FL  VN 

And  (125),    AK=:^=(38  and  99)    ^^     . 


AB 


By  substituting  these  values  into  the  proportion  above,  we  read- 
ily obtain 

Maj.FM.MV  :  Aw).FL.VN  ::  MG^  :  AO^. 

But  again,  since  in  the  limit  LM  coincides  with  MG,  and  AT 
with  AO,  FL  becomes  the  altitude  of  the  triangle  GFM,  and  AF  ot 
FAO,  and  the  areas  of  these  triangles  being  equal,  we  have* 

AF.AO=FL.MG. 
Or,  AF    :  FL    ::  MG    :  AO. 

And  squaring,     AF^  :  FL^  ::  MG^  :  AOl 
Hence,  by  equality  of  ratios  and  dividing  by  FL, 
Ma;.FM.MV  :  Aw^.VN  ::  AF^  :  FL. 


•  Leg.  4.  6. 

»  L.,4,  6;  W.,4,  5. 


OF  CURVATURE. 


76 


By  the  similar  triangles,  MNV  and  MLF  (30  and  92),  we  have 
the  proportion, 

•    MV  :  VN  ::  FM  :  PL. 

By  which  divide  the  preceding  one,  and  we  get 

AP 
Ma:.FM  :  Aw    ::  ^  :  I, 
FM 

Dividing  the  first  and  third  by  FM,  and  multiplying  the  third  and 
fourth  by  FM^, 

Mx  :  Aw   ::  AF^  :  FW. 


(Fig.  62— Parabola.) 


(134a)  Secondly,  in  the  parabola  it  may  be  shown,  in  the  sanw 
manner  as  in  the  ellipse  and  hyperbola,  that 

MxMR  :  Aw.AK  ::  MG^  :  AO^ 

And  also  that  AP  :  FL^  ::  MG^  :  A0«. 
Therefore,  by  equality  of  ratios, 

MxMR  :  Aw.AK  ::  AP  :  FL«. 


76  CONIC  SECTIONS. 

But  (129)  MR=4FM,  and  (125  and  69)  AK=«4AF,  and  (68c) 
FL2=AF.FM.     Hence,  by  substitution, 

Ma;.FM  :  Aw.AF  ::  AP  :  AF.FM. 

Dividing  the  second  and  fourth  by  AF,  and  transferring  the 
factor  FM  from  the  first  to  the  fourth, 

Mx  :  Aw  ::  AF^  :  FMl 

(135)  Schol  On  the  property  proved  in  this  proposition  depends 
the  important  law  tiiat  the  paths  of  the  heavenly  bodies,  and  of  all 
others  under  the  influence  of  gravitation,  are  necessarily  conic 
sections. 


OF  QUADRATURE. 


77 


CHAPTER  YI. 

OF  SOME  PROPERTIES   PECULIAR  TO  THE  DIFFERENT 
CONIC  SECTIONS. 


(136)  Prop.  I.     Theorem. 

T%e  area  of  a  parabola  is  equal  to  two-thirds  of  the  circumscribing 

parallelogram. 


That  is, 


MAP=fMBLP. 


Through  the  point  S  indefinitely  near 
to  M  draw  NF  parallel  to  the  abscissa 
AD,  and  RC  parallel  to  the  ordinate 
MD  ;  also  draw  the  normal  MG.  The 
triangles  MNS  and  MDG  are  similar," 
since  the  sides  of  the  one  are  respec- 
tively perpendicular  to  those  of  the  oth- 
er.    Therefore 

MN  :  SN  ::  DG  :  MD. 

Or,  multiplying  the  first  and  third  terms 
by  AD=MB,  and  the  second  and  fourth 
by  MD, 

MN.MB  :  SN.MD  ::  AD.DG  :  MD2=(60and686)2AD.DG  ::  1  :  2 

That  is,  the  interior  rectangle  MC  is  double  the  external  one 
MF,  and  the  same  would  be  true  of  any  other  rectangles  similarly 
drawn.  Consequently  the  whole  space  AMD  is  double  of  ABM, 
and  hence  equal  to  JABMD ;  or  MAP=|MBLP. 


•  Leg.  4.  21. 

•  L.,  4,  22  ;  W.,  3,  7  and  8. 


78 


CONIC  SECTIONS. 


(137)  Prop.  II.     Theorem. 

The  area  of  an  ellipse  is  to  that  of  a  circle  described  on  its  tranS' 
verse  axis,  as  the  conjugate  axis  is  to  the  transverse. 

That  is,     ADBE  :  AD"'BE'"  ::  DE  :  AB. 


Draw  any  two  ordinates  H'"F  and 
S'"G,  indefinitely  near  each  other. 

It  follows  readily  from  (21  )^  that 
the  area  of  HFGS  :  the  area  of 
H'"FGS'"  ::  CD  :  CD'"  ::  DE  : 
AB,  and  the  same  would  be  true  of 
any  other  trapezoids  similarly  drawn. 
We  may  therefore  suppose  them  in- 
definitely increased  so  as  to  occupy 
the  entire  area  of  the  ellipse  and 
circle,  and  shall  then  have  the  area 
of  ADBE  :  the  area  of  AD'"BE'"  :: 


(Fig.  64.) 


DE  :  AB. 


(138)  Prop.  III.     Theorem. 

TVie  sum  of  the  first,  second,  or  third  powers  of  four  lines  drawn 
from  one  of  the  foci  of  an  ellipse  to  the  extremities  of  any  pair 
of  conjugate  diameters  is  the  same,  whatever  may  be  the  position 
of  those  diameters. 

That  is,     VI  +VD  +VH  +VP  =a  constant  quantity. 
Or,  VF+VD2+VH2+VP2=a  constant  quantity. 

Or,  VP+VD3+VH3+VP3=a  constant  quantity. 

Join  FP  and  FL 

Put  the  semi-transverse  axis = a 

"    the  semi- conjugate  axis=6. 

"    the  eccentricity  (14) =c. 

"    Yl=x,  VH  or  FI=y,  VD  or  FP=a;',  and  VP=y'. 

*  Leg.  4.  7. 

•L.,  4,  7;  W.,  4,6. 


MEAN  VALUES.  ?• 

(Fig.  66.) 


Case  1. 

By  (28)        x-\-y=2a,  and  also  x'-]ry'=2a. 
Therefore    x-\-y-\-x'-\-y'=^Aa,  a  constant  quantity. 

(139)  Cor.  1.  Hence  the  mean  value  of  lines  drawn  from  one 
of  the  foci  to  different  points  of  the  curve  is  equal  to  the  semi- 
transverse  axis. 

Case  2. 

By  algebra  we  have     a:^+i/^=(a;+y)*— 2a;y. 

But  (28)     {x-\'yy=4.a\  and  (45)  2a:y=2CP". 

Therefore  x^^-y'^=^4:a^-2CV^. 

In  like  manner     a;'2+y'2=4a^— 2CP. 

Therefore  a:2+3/2+a;'2_|.y/2^8a2_2(CP+CI»)=(42)  8a«-2  {a^^l^ 
*=6a^—2b^={27)  4a^+2e\  a  constant  quantity. 

(140)  Cor.  2.  Hence  the  mean  value  of  the  squares  of  lines 
drawn  from  one  of  the  foci  of  an  ellipse  to  different  points  of  the 
curve  is  equal  to  a^-\-^(^ ;  that  is,  to  the  square  of  the  semi- transverse 
axis,  plus  half  the  square  of  the  eccentricity. 


fiO  CONIC  SECTIONS. 

Case  3 
By  algebra  we  have 

^-\-y^=(x-hyy—Sx^y-'Sxi/^=(x-^yYS  (x+y)  xy. 
But  (28)    x-\-y=^2a,  and  hence  {x-{-yy=Sa^;  also  (46)  a;y=CP*. 
Therefore  x^+f=8d^-6a,CF^. 

In  like  manner    x'^-{-y'^=8a^—6a.CP. 

Therefore  x^-^-y^-^-x'^-^y'^^^lSa^-Qa  (CP+CI2)=(42)  IGflS— 
6a  (a2+62)  =  i0a3_6afe2  =  4a8_^6a  (a2-fca)=4a8^6^c2,  a  constant 
quantity. 

(141)  Cor.  3.  Hence  the  mean  value  of  the  cubes  of  lines  drawn 
from  one  of  the  foci  of  an  ellipse  to  different  points  of  the  curve  is 
equal  to  a^+l^ae^;  that  is,  to  the  cube  of  the  semi- trans  verse  axis 
plus  the  square  of  the  eccentricity  multiplied  by  three-fourths  of  the 
transverse  axis. 

(142)  Prop.  IV.     Problem. 
\ 
To  find  the  mean  value  of  the  reciprocals  of  a  given  power  of  lines 
drawn  from  one  of  the  foci  of  an  ellipse  of  small  eccentricity  to 
different  points  of  the  curve. 

Put  the  semi-transverse  axis=a. 

"  the  semi-conjugate  axis =6. 

"  the  eccentricity=c. 

"  VI  (Fig.  55)=a+x. 

«  VH  or  FI=(28)  fl-a!. 

«  VP=a+?/. 

"  VD  or  FP=(28)  a—y. 

"  the  index  of  the  given  power=«. 

n  (7i+l>  (n-{-2)  x^ 

\.2.S.a^        '  ^'^' 


And     y^ 


OF  31EAN  VALUES.  81 


n  (71+1)  (71+2)  :^:»  ^^ 
1.2.3.a«+3 


By  the  conditions  of  the  proposition  the  value  of  a;  or  y  is  but  a 
small  fraction  of  a,  and  hence  these  series  converge  so  rapidly  that 
It  will  be  sufficiently  accurate  to  employ  only  the  first  four  terms. 
By  adding  the  two  series  together  we  obtain, 

1     .      1    ^  '^    ,  ^  (^+1)  ^ 


VI«^VH«    a"  a«+2 


r      |.|  1,1  2  71   (71+1)  f 

In  like  manner,    ,^^+,^^=-+ --^ . 

Hence   Jj,  +  ^„  +  ^„+^4+!Lgi^ 

mean  value  of  the  four  is 1 — f — tt^  (^^+y)- 

a"        4a"+'' 

Now  (45)      CP2=VLIF=(a+2:)  (a-a:)=a*-a:*. 
And,  in  like  manner,     CH^=fl^— y^. 
Therefore     CP2+CH2=2a2_(a;2+y2). 
But  (42)        CP+CH2=aa+6a,    and  (27)  l^=a^-(?. 
Therefoie     2a^-{3^-^f)==2a^-(^,    and  x«+y2=c2. 
By  substituting  c^  in  the  place  of  (a;^+i/^  in  the  foregoing  ex- 
pression for  the  mean  value,  it  becomes 1 — ^ — -ri — ,  an  expres- 

sion  containing  only  constant  quantities ;  and  since  HI  and  DP  are 
any  conjugate  diameters,  it  must  be  true  for  the  entire  circum- 
ference. 

(143)  Schol.  This  proposition  enables  us  to  find  the  mean  at- 
traction of  the  sun  upon  any  planet  throughout  its  entire  orbit,  and 
would  do  so  equally  well  if  the  force  of  gravity  varied  inversely  as 
the  third,  fourth,  or  any  higher  power  of  the  distance.*    By  means 

•  That  18,  if  we  suppose  it  possible  for  the  law  of  gravity  to  be  changed,  and  the 
orbit  still  to  ret-ain  its  elliptical  form. 

II 


82  CONIC  SECTIONS. 

of  the  principle  involved  in  it  Laplace  succeeded  in  discovering  the 
true  cause  of  the  secular  acceleration  of  the  moon's  mean  motion^  a 
subject  w^hich  had  very  much  perplexed  previous  astronomers. 

In  the  case  of  gravity  7i=2,  and  the  expression  for  the  mean 

1       3c^ 
value  becomes  —5  +  ;r-i,  from  v^hich  we  learn  that  the  attraction  is 
a^      2a* 

greater  than  it  v^^ould  be  if  the  planet  revolved  in  a  circle  at  the 

same  mean  distance  in  the  ratio  a*-\-\\(?  :  a* 


PART   II. 


ANALYTICAL  &EOMETRY. 


CHAPTER  I. 

OP   THE   POINT   AND   STRAIGHT   LINE   IN   A  PLANE. 


(Fig.  67.) 

Y 


(144)  The  position  of  a  point  in  a  plane  may  be  determined  in 
either  of  two  ways,  viz. :  by  determining  its  distances  from  two 
given  lines  in  the  plane  that  intersect  one  another,  or  by  deter- 
mining its  distance 
and  direction  from  a 
given  point  in  the 
plane.  Thus,  the  po- 
sition of  the  point  P 
may  be  determined 
by  knowing  its  dis- 
tances PD  and  PE 
from  the  fixed  lines  x'. 
AX  and  AY,  in  which 
case  it  is  said  to  be 
determined  by  rectili- 
neal co-ordinates ;  or 
by  knowing  the  dis- 
tance AP  and  the 
angle  PAD,  in  which 
case  it  is  said  to  be  determined  by  polar  co-ordinates.  Hence,  in  a 
plane,  rectilineal  co-ordinates  consist  of  two  straight  lines,  and  polar 
co-ordinates  of  a  line  and  an  angle.  .»    j^ 


E 

I 

> 

A 

^                          1 

) 

Y 

J 


84  ANALYTICAL  GEOMETRY. 

(145)  In  the  former  case  the  fixed  lines  X'AX  and  YAY'  are 
called  co-ordinate  axesy  or  axes  of  reference,  and  taken  separately 
the  first  is  called  the  axis  of  abscissas,  and  the  second  the  axis  of 
tirdinates.     The  point  of  intersection  A  is  called  the  origin. 

(146)  The  line  PE  parallel  to  AX  is  called  the  abscissa  of  the 
point  P,  and  the  line  PD  parallel  to  AY  is  called  the  ordinate. 
Taken  together  they  are  called  co-ordinates,  as  already  remarked. 
Instead  of  PE  we  may  employ  its  equal  AD  as  the  abscissa. 

(147)  If  the  axes  are  at  right  angles  to  one  another  the  co-ordi- 
nates are  said  to  be  rectangular,  but  if  not  they  are  called  oblique. 

(148)  The  angle  YAX  is  called  the  first  angle,  YAX'  the  second, 
X'AY'  the  third,  and  Y'AX  the  fourth. 

(149)  All  ordinates  drawn  upwards  from"  X'AX  are  considered 
positive,  and  those  drawn  downwards  negative ;  while  all  abscissas 
drawn  from  YAY'  to  the  right  are  considered  positive,  and  those 
drawn  to  the  left  negative.  Hence  the  co-ordinates  of  a  point 
situated  in  the  first  angle  are  both  positive ;  in  the  second,  the  ab- 
scissa is  negative  and  the  ordinate  positive ;  in  the  third,  they  are 
both  negative ;  and  in  the  fourth,  the  abscissa  is  positive,  but  the 
ordinate  is  negative. 

(150)  It  is  plain  that  a  single  point  can  have  but  one  abscissa, 
and  but  one  ordinate;  but  a  line,  since  it  contains  an  indefinite 
Qumber  of  points,  can  have  an  indefinite  number  of  pairs  of  co- 
ordinates, varying  in  their  length,  and  hence  spoken  of  as  variable 
quantities.  It  is  customary  to  denote  the  abscissa  by  the  letter  x, 
and  the  ordinate  by  y. 

(151)  If  polar  co-ordinates  are  employed  the  point  A  is  called  the 
vole,  the  line  AX  the  angular  axis,  AP  the  radius  vector,  and  PAX 
the  variable  angle. 

In  pursuing  the  subject  we  shall  at  first  employ  rectangular  co- 
ordinates, as  these  are  more  simple  in  their  application ;  but  shall 


OF  STHAlciHT  LINES. 


86 


now,  before  closing  the  chapter,  how  they  may  be  transformed  into 
jblique  or  polar  ones. 

(152)  Def.  The  equation  of  a  line,  whether  straight  or  curved^ 
IS  one  that  expresses  the  relation  between  the  co-ordinates  of  any 
point  in  the  line. 

For  example,  if  in  Fig.  57  PD  is  two-thirds  £is  long  as  PE  or 
AD,  the  same  ratio  would  exist*  between  the  co-ordinates  of  any 
other  point  in  the  line  AP ;  so  that  wherever  they  were  drawn  we 
should  have  y=fa;.     This  is,  therefore,  the  equation  of  the  line  AP 


(153)  Prop.  I.     Theorem. 

The  general  equation  of  a  straight  line  is 

y  =  ax-\-h; 
in  which'  a  represents  the  tangent  of  the  angle  that  the  line 
makes  with  the  axis  of  abscissas,  h  the  portion  of  the  axis  of  or- 
dinates  intercepted  between  the  line  and  the  origin,  and  j?  and  y 
the  co-ordinates  of  any  point  in  the  line. 

Let  AX  and  AY  be  the  axes,  A  the  origin,  PT  any  straight  line, 
and    AD   and    PD    co-ordi- 
nates of  any  point  P  in  the 
line. 

Put  AD=a:. 

"     PD=!/. 

"    AR=6. 

"    tan.  PTD=a. 


(Fig 

Y 

.  68.) 

1 

?^^ 

R 

F 

, 

] 

X 

By  trigonometry,  PF=RF  tan.  PRF. 
But^  tan.  PRF=tan.  PTD=a, 

Therefore     PF=aa:. 
Also  FD=AR=i?>. 

Therefore     aa;+6=PF+FD=PD=v. 
Or  y=ax-\-h. 


and  RF=AD=a:. 


•  Ix-g.  4.  18.     Euc.  6.  4. 
-  L.,4,  19;  W.,  3,4,  and  5. 


'  Leg.  1.  20,  Cot.  3.    Euc.  1.  39. 
*•  L    1,22;  W.,  1,  12,  cind  14. 


86 


ANALYTICAL  GEOMETRY. 


(154)  Cor.  1.  If  the  line  passes  through  the  origin,  /)=o,  and  the 
equation  reduces  to  y—ax,  a  form  analogous  to  that  used  above  as 
an  example,  in  which  a—\. 

(155)  Cor.  2.  If  the  line  is  parallel  to  the  axis  of  abscissas,  a— 
o,  and  the  equation  reduces  to  y=b.  It  is  evident  also,  from  an 
inspection  of  the  figure,  that  in  that  case  every  ordinate  v^rould  be 
equal  to  AR. 


(156)  Prop.  II.     Theorem. 

The  equation  of  a  straight  line  passing  through  a  given  point  is 
y'^y=a  {x'—x); 
in  which  x'  and  y'  represent  the  co-ordinates  of  the  given  point, 
X  and  y  those  of  any  other  point  in  the  line,  and  a  the  tangent 
of  the  angle  that  the  line  makes  with  the  axis  of  absdissas. 

Let  RT  be  the  straight  line,  M  the  given  point,  and  AD  and  PD 
the  co-ordinates  of  any 
other  point  P  in  the  line.  ^^'^^'  ^^'^ 

Put  AD=a;. 
"    PD=y. 

"    AN=a;'.  T 

"    tan.  PTD=tan.  MPE=a. 

Then     ME=y'-y,     and  PE=a;'-a?. 
But       ME=PE.  tan.  MPE. 
Therefore     y'—y^=a  (x' — x). 

(156a)  Otherwise;  since  Prop.  I.  is  true  of  every  point   n  TM< 
we  have  the  two  equations, 

y  ~ax  ■{■}), 
\  y'=ax'-\'b. 

Eliminating  b  by  means  of  these  equations,  we  have 
y'-^y=a  (x'—x). 


OF  STRAIGHT  LINES. 


87 


(157)  Prop.  III.     Theorem. 
The  equation  of  a  straight  line  passing  through  two  given  points  u 

in  which  x'  and  y'  represent  the  co-ordinates  of  one  of  the  given 
points,  x"  and  y"  those  of  the  other,  and  x  and  y  those  of  s 
other  point  in  the  hne. 

Let  RT  be  the  straight  line,  M  and  R  the  given  points,  and  AD 
and  PD  the  co-ordinates  of  any  point  P. 

Put  AD  =x. 

^^  (Fig.  60.) 

"  AN=a:'. 

"  MN=y'. 

"  AS  =x", 

"  RS  =y''. 

Then    ME=y'-y,  PE=x'-x,  RF=y"-y',  and  MF=:r"— r 

By  sim.  tri.     MF  :  RF  ::  PE  :  ME. 

Hence 

That  is,  yr^y=lf:^(x^^x). 


MF.ME=RF.PE,    or,  ME  =  ^PE. 


x"-x' 

(157a)  Otherwise ;  tan.  PTS=  tan.  RMF=(by  trig.)  ^^ -„_,' 
which  substitute  in  the  place  of  a  in  the  last  proposition,  and  we 

get,  as  before,  y'-^y=z^       "^  (x'—x). 
X  —^x 

(1576)  Otherwise;  by  Prop.  I.  we  have  the  three  equations, 

y  —ax  +6. 

y'  =ax'  +6, 

y"=ax"^-h. 
Eliminating  a  and  b  by  means  of  these  three  equations,  we  have 


68 


ANALYTICAL  GEOMETRY. 


(158)  Prop.  IV.     Theorem. 

Every  equation  of  the  first  degree  between  two  variables  is  the  equa* 

tion  of  a  straight  line. 

For,  by  transposing  and  uniting  terms,  every  such  equation  can 
be  reduced  to  the  form  A?/+Ba:+C=0,  in  which  A,  B,  and  C  repre- 
sent any  constant  quantities,  whether  positive  or  negative.    But  the 

B      C    .       ^.  ^      B 

above  equation  reduces  to  y~~T^~T>  ^^  which  — r-  answers  to 

C 
a  in  the  first  proposition,  and  — ^  ^^  ^ »  that  is,  it  reduces  to  the 

form  y=ax-)rb. 

Example.     Prove  that  the  equation  20— a;+7y=10a:— 12  is  the 

equation  of  a  straight  Hne. 


(159)  Prop.  V.    Theorem. 
ITie  equation  of  the  distance  between  ttvo  points  is 

d  =  vW^yY  +  i^'-^Y; 

in  which  x'  and  y'  are  the  co-ordinates  of  one  point,  x  and  y 
those  of  the  other,  and  d  the  distance  between  the  two  points. 

Let  M  and  R  be  the  two  points.  y         (Fig.  61.) 

Put  AN=a:. 

"    MN=y. 
"    AS  =x\ 
"    RS=y. 

Then  RF=y  -y,     and  MF=«a:'— a:. 

Buf^    MR*=RF"+MP=(y'-y)'+(a:'-a?)«. 

And,  extracting  the  root,    MR='^(y'— y)2+(a;'— a:)*. 


"  Leg.  4.  11.    Euc.  1.  47. 
.L.,4,11;  W.,4,8. 


OF  STEAIGHT  LINES. 


89 


(160)  Prop.  VI.     Theorem. 

The  equation  of  the  tangent  of  the  angle  included  between  two  straight 

lines  is 

_   a'  —  a 

in  which  a  represents  the  tangent  of  the  angle  that  one  of  them 
makes  with  the  axis  of  abscissas,  and  a'  the  other,  and  t  the  tan- 
gent of  the  angle  formed  by  the  lines. 

Let  VR  and  VS  be  the  two  lines  (Fig.  ea.) 

intersecting  one  another  in  V. 

Put  tan.  VRX=a. 
"    tan.  VSX=a'. 

Because  VSX  is  the  exterior  angle 
of  the  triangle  VRS,  we  have''  VSX= 
RVS+VRS,  or  RVS=VSX-VRX; 
and  consequently,  tan.  RVS  =  tan. 
(VSX-VRX). 

But,  by  trigonometry,''  the  formula  for  the  tangent  of  the  difler 

.     a' — a 
ence  of  two  angles  whose  tangents  are  a'  and  a,  is  — - — j. 

a'— a 


Therefore    tan.  RVS=- 


1+aa' 


(161)  Cor,     If  the  Hues  are  parallel  a'  —  a—o,  and  il  perpendic 
ular  to  one  another  \-{-aa'  =  o.     For  when  the  value  of  a  fraction 
is  nothing  its  numerator  must  be  nothing,  and  when  the  value  is 
infinite  the  denominator  must  be  nothing. 

We  have  thus  far  used  only  rectangular  co-ordinates,  but  it  is 
sometimes  more  convenient  to  employ  oblique  or  polar  ones.  It 
is,  therefore,  important  to  be  able  to  pass  from  one  system  to  the 
other ;  that  is,  to  be  able  to  find  the  oblique  or  polar  co-oidinates 


Leg.  1.  26,  Cor.  6.    Euc.  1.  32.  *  Leg.  Trigonometry,  Art  XXV. 

L.,  1,  27.  »»  Loomis's  Trig.,  Art.  76. 


90 


Ax\ALYTICAL  GEOMETRY. 


of  a  point  from  the  rectangular  ones,  and  the  reverse.     The  process 
is  called 


ALH 


TRANSFORMATION    OF    CO-ORDINATES 
(162)  Prop.  VII.     Problem. 

To  pass  from  a  system  of  rectangular  to  a  system  of  oblique  co 

ordinates. 

Let  AX  and  AY 
be  the  primitive  rect- 
angular axes,  and  AF 
and  FP  the  co-ordi- 
nates of  any  pomt  P. 
Let  A'X'  and  A'Y' 
be  the  new  axes,  A'S 
and  PS  the  new  co- 
ordinates of  the  point 
P.     Also,  let  AH  and 

A'H  be  the  co-ordinates  of  the  origin  of  the  new  system,  and  X'GX 
and  Y'LX  the  angles  which  the  new  axes  make  with  the  primitive 
axis  of  abscissas. 

Put  AF=a:,  PF=2/,  A'S=z',  SP=y',  AH=7n,  A'H=w,  X'GX 
«=a,  and  Y'LX=a'. 

Then,  by  trigonometry, 

MS=A'S,  cos.  A'SM=a:'  cos.  a. 
And     SD  =  SP.  cos.  PSD  =y'  cos.  a'. 

Therefore    x'  cos.  a+y'  cos.  a'=MS  +  SD  =  HF. 

(162«)  Or,  adding  AH,  x'  cos.  a+y'  cos.  a'+m=AF=a:. 

Also,  by  trigonometry,  A'M=A'S.  sin.  A'SM=a;'  sin.  a. 

And,  adding  A'H,  MH  or  DF=x'  sin.  a+n. 

Also,  by  trigonometry,  PD  =  SP,  sin.  PSD=y'  sin.  a'. 

(1626)  Hence     x'  sin.  a-^-n+y'  sin.  a'=DF-f-PD=PF=v. 


OF  STRAIGHT  LINES.  91 

Now,  if  in  any  equation  of  a  line  referred  to  the  primitive  axes 
we  substitute  for  x  and  y  their  values  just  obtained,  we  shall  have 
the  equation  for  the  new  axes. 

(163)  Cor.  1.  If  the  origin  is  the  same  in  both  systems,  m  and 
n  disappear,  and  the  expressions  for  x  and  y  become 

x=x'  cos.  a-\-y'  cos.  a',  and  y  =  x'  sin.  a-\-y'  sin.  a'. 

(164)  Cor.  2.  If  the  new  axes  are  parallel  to  the  primitive  ones, 
<i=0°,  and  a'=90°,  which  reduces  the  expression  for  x  to  x'-\-m, 
and  for  y  to  y'-\-n. 

(165)  Cor.  3.  If  the  new  axes  are  rectangular,  but  not  parallel 
to  the  primitive  ones,  «'=90°+a,  which  changes  the  expression  for 
X  to  x'  cos.  a  —  y'  sin.  a-\-m,  and  that  for  y  to  x'  sin.  a-\-y'  cos.  a  +  n 

For  then,'^     sin.  «'=sin.  90°  cos.  a  +  cos.  90°  sin.  a=cos.  a. 
And  cos.  rt'  =  cos.  90°  cos.  a— sin.  90°  sin.  a=  — sin.  a. 

(166)  Schol.  It  matters  not  in  which  of  the  four  angles  formed 
by  the  primitive  axes  the  origin  of  the  new  ones  is  placed,  provided 
the  proper  signs  are  prefixed  to  the  co-ordinates  m  and  n.  And 
further,  since  the  only  effect  of  these  co-ordinates  is  to  add  their 
lengths  to  the  values  otherwise  obtained,  we  may,  in  any  case,  first 
find  the  value  of  the  primitive  co-ordinates  in  terms  of  the  new 
ones,  and  then  add  the  co-ordinates  of  the  new  origin. 


(167)  Prop.  VIII.     Problem. 

To  pass  from  a  system  of  oblique  to  a  system  of  rectangular  co^ 

ordinates. 

Using  the  same  notation  and  figure  as  in  th    lust  proposition,  and 
supposing  the  origin  to  be  the  same  in  both  s}  .tems",  we  have  (163) 

x^x'  cos.  a+y  COS.  a\   and  y=x'  si  a.  a+y'  sm.  a'. 


'  Davies'  Legendre,  Trigonometry,  Art.  XIX. 


92 


ANALYTICAL  GEOMETRY. 


Solving  these  equations  for  a:'  and  y'j  and  adding  (166)  ihe  co- 
ordinates of  the  new  origin,  designated  by  m'  and  n\  we  obtain 

X  sin.  a'—y  cos.  a'  ,      ,      ^  sin.  a' —y  cos.  a'  , 

sm.  a  cos.  a — sin.  a  cos.  a'  sin.  \a' — a) 

y  cos.  a—x  sin.  a 


y-- 


.     ,      V  cos.  a— a;  sin.  a  ,     , 

7  +  71'  =  ^ -. — r \-n'. 

sm.  a'  cos.  « — sin.  a  cos.  a'  sin.  (a  —a) 


Now  if,  in  any  equation  of  a  line  referred  to  the  primitive  ob- 
lique axes,  we  substitute  for  x'  and  y'  their  values  just  obtained,  we 
shall  have  the  equation  for  the  new  rectangular  axes. 

(168)  Schol.  By  the  aid  of  the  two  preceding  propositions  we 
can  pass  from  one  system  of  oblique  co-ordinates  to  another,  by 
first  passing  from  the  primitive  oblique  system  to  a  rectangular 
one  by  Prop.  VIIL,  and  from  that  to  the  new  oblique  system  by 
Prop.  VII. 


(169)  Prop.  IX.     Problem. 

Vo  pass  from  a  system  of  rectangular  to  a  system  of  polar  Oh 

ordinates. 

Let  AX  and  AY  be  the  (Pig.  66.) 

primitive  axes,  and  AF  and 
FP  the  co-ordinates  of  the 
point  P. 

Let  A'X'  be  the  new  angu- 
lar axis,  A'  the  pole,  AH  and 
A'H  its  co-ordinates,  A'P  the 
radius  vector,  PA'X'  the  va- 
riable angle  (151),  and  DA'X' 

the  angle  which  the  new  angular  axis  makes  with  the  primitive 
axis  of  abscissas. 

Put  AH=m,  A'H=n,  A'P=r,  DA'X'=a,  X'AT=«,  AF=a, 
and  PF=y. 


'  Davies'  Legendre,  Trigonometiy,  Art.  XIX. 


OF  STRAIGHT  LINES.  93 

Then  HF=A^D= (by  trigonometry)  AT  cos.PA'D=r  cos.  (a+w) 

(169a)  And  AF=AH+HF=m  +  r  cos.  (a+w)=a:. 
Again,  by  trigonometry,     PD=A'P  sin.  PA'D=r  sin.  (a+w). 

(1696)  And  PF=A'H+PD=7i+r  sin.  {a-\-ui)=y. 

These  values  of  x  and  y  being  substituted  in  any  equation  in  ttie 
same  manner  as  in  Prop.  VII. ,  will  give  us  the  polar  equation. 

(160c)  Schol.  By  substituting,  in  place  of  sin.  (a  +  (i>)aiid  cos.  (a-f-w), 
their  values  in  terms  of  these  angles  separately,^  and  solving  the  resulting 
equations  for  r  and  w,  we  shall  find  equations  for  passing  frona  a  system  of 
polar  to  a  system  of  rectangular  co-ordinates. 

•  Daviea'  Legendre,  Trigonometry,  Art  XTX. 


fM 


ANALYTICAL  GEOMETRY 


CHAPTER     II, 


OF  CURVES 


(170)  Prop.  I.     Problem. 

^  To  construct  a  curve  from  its  equation. 

Draw  the  axes  X'AX  and  AY ;  in  X'AX  take  at  pleasure  anj 
points,  —1,  1,  2,  3,  4,  5,  (fee,  and  calculate  the  corresponding  ordi- 
nates  from  the  equa- 
tion   of    the    curve.  ^        ^^- ^^'^ 
Lay  off  these   ordi- 
nates  parallel  to  AY, 
and  above  or  below 
X'AX,  according  as 
the    value    of    y    is 
found    to     be    posi- 
tive or  negative,  and 
through  their  extrem- 
ities, as  fl,  h,  c,  d,  (fee,  draw  the  curve. 

For  example,  suppose  the  equation  of  the  curve  to  be  y—aa^, 
and  that  a  =  5.  Giving  arbitrary  numerical  values  to  x,  (suppose 
the  series  of  numbers  from  —4  to  +5,  as  shown  below,)  and  from 
the  equation  computing  for  each  value  of  x  the  corresponding  value 
of  y,  we  obtain  the  series  set  against  y. 


X'- 


— 1 


x=     -4,       -3,     -2,   -1,  0,  1,     2,       3,       4,       5, 
y  =— 320,   -135,   —40,   -5,  0,  5,  40,  135,  320,  625. 

These  values,  being  laid  off  upon  the  axes  according  to  theit 
signs  (149),  will  enable  us  to  trace  the  curve. 


OF  CURVES. 


95 


(171)  Examples,  Construct  the  following  cun  es,  putting  a=8, 
0=5, 77i=10,  n=8,  p=ly  and  r=6;  and  taking  care  in  the  last  six 
to  draw  both  values  of  y.  ' 

1st.     TTiy  =  aa;2+ 6,  giving  to  x  values  from   —5  to  +5  inclusive 

2d.        y=x^-{-h,         "  "  "         —5  to   +5 

8d.        y=bx-a,         "  "  "         -5  to   +5 

4th.      {y—ny=T^—(x—my,  giving  to  x  values  from  +3  to  +17 

mclusive. 

5th.     y^=ao(^—a^+b,  giving  to  x  values  from  —4  to  +10  inclusive, 

and  also  the  value  8,07. 
6th.     y'^=7^—a^y  (riving  to  x  values  from  —7  to  +7  inclusive. 
7th.     y^=px,  "  "  "     —2  to +8 


«th.     y'=5(«'-^    " 
9th.     y^=^^(a^^a')    « 


«     -9  to  +9 

"  —16  to  +16       ** 


(172)  Prop.  II.     Theorem. 

The  equation  of  a  circle  is  (x— 77i)^+(y— n)*— r*=0;  in  which  m 
and  n  denote  the  co-ordinates  of  the  centre,  and  r  the  radius. 

Let  PNP'  be  a  circle,  AB  and  BC  (Fig-  67.) 

the  co-ordinates  of  the  centre  C,  and 
AE  and  EP  those  of  any  point  P  in 
the  circumference. 

Draw  CD  parallel  to  AX,  and  join 
CP.  Put  AB=m,  BC=7i,  CP=r,  AE 
=07,  and  EP— y. 

Then  CD==AE— AB  =  a:— 772,  and 
PD=i;P-BC=^ite=y:  /^^  W 

Bur  CD3+PD2  =  CP«. 
That  is,     {x—mf+(y-ny=f*; 
or,  by  transposition,     (x — m)^  +  (y — n)^  —  r" = 0. 


96 


ANALYTICAL    GEOMETRY. 


(173)  Cor.    If  the  origin  is  at  the  centre,  m  and  n  disappear,  and 
the  equation  reduces  to  x^-\-y'^—T^  =  ^,  a  form  often  met  with. 

(174)  Prop.  III.    Problem.    {^See  page  \m.) 

To  find  the  equation  of  a  tangent  to  a  circkf  the  origin  of  the  co 
ordinates  being  the  centre, 

(Fig.  68.) 


Let  MT  be  a  line  touching  the  circle  BMP  at  the  point  M,  it  is 
required  to  find  its  equation. 

Since  it  passes  through  the  point  M  its  equation  must  (156)  be 
of  the  form  y'—y=a  (x'—x),  in  which  x'  and  y'  are  the  co-ordi- 
nates of  the  point  M,  x  and  y  those  of  any  other  point  H  in  the 
line  MT,  and  a  the  tabular  tangent  of  the  angle  MTX,  or  its  equal 
•AMN. 

R  :  tan.  AMN. 

1    :  a 


By  trig. 

MN  : 

AN 

That  is, 

y'     : 

z' 

Hen^e 

x' 

(175)  Now  of  the  three  values,  a,  x'  and  y',  that  enter  into  tms 
equation,  either  all  are  negative,  or  one  only.  Thus,  if  the  point  be 
taken  in  the  first  angle,  as  at  M',  a:'  and  y'  will  be  positive  (149), 


•  Leff.  4.  23      Euc.  6,  8. 

*  L.,  4,  23,  Cor.    W.,  2,  14,  Cor.  1. 


OF  CURVES.  97 

but  the  tangent  of  M'T'X  negative,  since  the  angle  is  between  90** 
and  180°.*  If  it  be  taken  in  the  second  angle,  as  at  M,  oc?  will  be 
negative,  y'  positive,  and  a  positive,  since  the  angle  MTX  is  be- 
tween 0°  and  OO**. 

In  the  same  manner  it  may  be  shown  that  if  the  point  be  taken 
in  the  third  quadrant,  a,  a?',  and  y'  will  all  be  negative ;  and  in  the 
fourth,  a  and  x'  positive,  but  y'  negative.  JSo  that  the  equation  will 
read — 

x' 
In  the  1st  quadrant     —  a=— ; 


y 


In  the  2d 
In  the  3d 


— a= 


-x' 


-y" 


In  the  4th       «  «=— ,. 

-y' 

.  All  of  which  can  be  reduced  to  the  form 
(175a)  a=~. 

y 

Substituting  this  value  of  a  into  the  equation  y'—y^a  (a;'— x) 
we  obtain  for  the  equation  of  the  tangent  MT, 

(1756)  y'^y^J'-(^:c'-x)', 

which  may  readily  be  reduced  to  the  form  x'x-\-y'y  —  i^  =  0,  by 
clearing  of  fractions,  transposing,  and  for  x'^-\-y'^  substituting  the 
equal  value  r*.     See  Appendix  B. 

•  It  is  proved  in  treatises  on  trigonometry  (see  Davies'  Legendre,  Trigonometry; 
Art.  Xn.)  that  the  tangent  of  an  angle  is  positive  when  the  angle  is  between  OC 
and  90O,  or  between  180^  and  270© ;  but  negative  when  the  angle  is  between  90C 
and  180O,  or  between  270o  and  360©. 

13 


^ 


as 


ANALYTICAL   iEOMETRY. 


(176)  Prop.  IV.     Problem 
To  find  the  polar  equation  of  a  circle 


(Fig.  69.) 


Let  P  be  the  pole,  and 
PX'  parallel  to  AX  the 
angular  axis. 

Put  the  radius  vector 
PM=^^  the  variable  an- 
gle MPX'  =  «,  AD  and 
PD  the  co-ordinates  *of 
the  pole=m  and  n,  and 
AN  and  MN  the  co-or- 
dinates of  M=a;  and  y. 

Then.    MN=MO+ON=MO+PD=(1696)  r'  sin.  w-fn  =y. 
And       AN=DN+AD=  PO-f-AD=(169«)  r'  cos.w+w=::r. 

Squaring  these  values  of  x  and  y,  substituting  them  into  the 
equation  of  the  circle  referred  to  its  centre,  which  is  (173)  o(^-\-y^— 
?'*  =  0,  and  recollecting  that  sin.^  w  +  cos.^  w  =  l,  we  obtain  the 
equation 

r'^-\-2  (m  COS.  w-f?i  sin.  w)  /-fm^  +  n^— r^  =  0, 

which  solved  for  r*  gives 


Y 

/ 

/i 

\ 

/ 

/py; 

\ 

[ 

//i 

" 

\ 

/ 
/ 

/   / 

M^ 

^ 

^   ■ 

r'=^  —  m  COS.  u—n  sin.  w  ±  "■^r^—rr^ — 7i^+(m  cos.  w-fn  sin.  w)^, 

which  is  the  equation  required,  the  two  values  of  r'  representing 
PMandPM'. 

Strictly,  however,  it  is  only  the  positive  value  of  r  that  we  are 
to  take  into  account,  for  PM'  is  not  truly  the  radius  vector,  but 
rather  a  continuation  of  it  backward  till  it  meets  the  curve  in 
another  point.  The  same  will  be  true  in  all  future  cases  when  the 
value  of  the  radius  vector  is  negative 

(177)  Schol.  If  the  pole  be  placed  on  the  line  AX,  it  is  evident 
that  all  the  terms  of  the  foregoing  equation  which  contain  n  will 


OF  CURVES. 


99 


disappear;  if  it  be  situated  on  AY,  all  that  contain  m  will  dis- 
appear ;  and  if  at  the  origin  A,  both  m  and  n  will  disappear,  which 
will  reduce  the  equation  in  the  latter  case  to 

r'=  ±r, 

the  polar  equation  when  the  pole  is  at  the  centre. 


(178)  Prop.  V.     Theorem. 

The  equation  of  an  ellipse  referred  to  its  axes  is  a^y^-{-b^oc^-  a^b^=0; 
in  which  a  and  b  are  the  semi-axes. 


(Fig.  70.) 

D 


Let  EBDA  be  the  ellipse,  AB 
and  DE  its  axes,  and  CG  and 
GS  the  co-ordinates  of  any  point 
S. 

Put  AC=a. 
"  CD=&. 
«  CG=x. 
«    SG=y. 

By  (24)  SG2  :  AG.GB  ::  CD*  :  ACl 

But  AG=AC-}-CG,  and  GB=AC-CG, 

and  consequently,  AG.GB=(AC+CG)  (AC-CG)=»AC>-CG« 

Therefore  SG^  :  AC^-CG^  ::  CD^  :  ACl 

That  is,  y2      .  ^2_^         ..  ^2      .  ^2 

Converting  this  proportion  into  an  equation,  we  have 

which  is  the  equation  of  an  ellipse,  and  can  readily  be  reduced  to 
either  of  the  following  forms,  viz. : 

(178a)  y^=^^{a^^x^,     or  aV-^lr'a^-a^b^^O. 


•  Leg.  4.  10.    Euc.  2.  5,  Cor. 

"  L.,4,  10;  W.,  §377. 


100 


ANALYTICAL  GEOMETRY. 


(1786)  SchoL  By  a  similar  process  we  can  obtain  from  the 
property  discussed  in  (54),  the  equation  of  an  ellipse  referred  to 
any  two  conjugate  diameters,  viz. : 

in  which  a'  and  b'  represent  the  semi-conjugate  diameters. 

(179)  Cor.  If  a=b,  the  ellipse  becomes  a  circle ;  for  then  the 
last  equation  will  reduce  to  y^+a;^— tf^=0,  which  is  (173)  the  equa- 
tion of  a  circle. 


(180)  Prop.  VI.     Problem. 
To  find  the  equation  of  a  tangent  to  an  ellipse. 

Let  MT  be  a  line  touching  the  ellipse  at  M,  it  is  required  to  fii  d 
its  equation. 

Since  it  passes  through 
the  point  M  its  equation 
must  (156)  be  of  the  form 
y'  —  y=a'  {x' — x),  in  which 
x'  and  y'  are  the  co-ordi- 
nates of  the  point  M,  x  and 
y  those  of  any  other  point 
H  in  the  line  MT,  and  a' 
the  tabular  tangent  of  the  angle  MTX. 

Put  AC=a,  and  CD =6. 

It  was  shown  (24)  that 

CD«  :  AC2  ::  MN^  :  AN.NB=(396)  CN.NT. 

Hence       NT  =  -^,-^=^. 


(Fig.  71.) 

V 

1 

M^,^ 

•^ 

/-if 

t 

s 

"\ 

G     Ia        I 

V 

^     "^ 

By 

Now,  by  trigonometry,     NT 
That  IS,     ^ 

which  reduced,  gives    a'  =  -^-;. 


MN  ::  R 
y       ::  1 


tan.  MTX. 


OF  CURVES. 


101 


It  may  be  shown  in  the  same  manner  as  in  the  circle  (175)  that 
of  the  three  quantities  a',  x',  and  y'  in  this  equation,  either  one  only 
or  all  three  must  in  every  case  be  negative,  while  h^  and  a*  are 
always  positive,  and  consequently  that  the  foregoing  equation  will 
in  all  cases  become 

(180a)  '^'  =  -^- 

Substituting  this  value  of  a'  into  the  equation  y'—y=a'  {x'—x) 
we  obtain  for  the  equation  of  the  tangent  MT, 


(180&) 


y'-y  = 


a^y< 


•X), 


This  can  be  reduced  to  a  more  simple  form  by  clearing  it  oi 
fractions,  and  subtracting  it  from  the  equation  of  the  ellipse,  viz.  ' 
a^y'^=a^y^—Wx^.     It  will  then  read,  after  transposition, 

(180c)  ahfy'-\-}^xx' -0^1^=0. 


(181)  Prop.  VII.     Problem. 
To  find  the  polar  equation  of  an  eihpse. 


Let  P  be  the  pole,  and 
PX'  parallel  to  AX  the  an- 
gular axis. 

Put  the  radius  vector  PM 
=r,  the  variable  angle  MPX' 
=  w,  AD  and  PD  the  co-or- 
dinates of  the  pole  -771  and 
n,  and  AN  and  NM  the  co- 
ordinates of  the  point  M=x 
and  y. 

Then,    as    in    the    circle 
(176),     a?=r  COS.  «+»i, 
and    y=r  sin.  w-4-n 


— X 


102  ANALYTICAL  GEOMETRY. 

Squaring  these  values  of  x  and  y,  and  substituting  them  into  the 
equation  of  the  elHpse  (178),  we  obtain 

(a2sin.»w+68cos.2a))r2+2(a27isin.w+&2^cos.w)r+rtV+6W-a2^^=0. 

which  solved  for  r  gives 

a^n  sin.  u-^b^m  cos.  w 


(181a) 


a^  sin.^  w  +  b^  cos.^  u 


/  a^b^—a^TV^—bV        /A  sin.  w  +  6^?7i  cos.  wV 
^  fl^sin.^w  +  t^cos.^w  ■*"  V  a^sin.^w  +  i^cos.^co  /  ' 

which  is  the  equation  required,  the  positive  value  of  r  representing 
PM,  and  the  negative  PM'. 


(182)  Schol.  1.  If  the  pole  be  placed  in  the  centre,  the  terms 
containing  m  and  n  disappear,  and  the  equation  reduces  to 

ab 

y.  =  -J-  z===^z=====:  ; 

v^a^  sin.^  w  +  6^  cos.^  w 
making  AM  equal  to  AM",  as  it  evidently  ought  to  be. 

(183)  SchoL  2.  If  the  pole  be  placed  at  F,  one  of  the  foci,  the 
polar  equation  may  be  obtained  more  easily  by  the  following 
process 

In  the  equation  of  the  ellipse  referred  to  rectangular  axes  (178«), 

viz.:  y^=-^(a^-x^y  substitute,  in  the  place  of  6^  its  value  (27) 

c?^m^,  and  it  will  read 

ft— •771 

Now  FM2=FN3+MN2=(a:-m)2+ —{a^-^a^^(by  reducing) 

_■     «       .  ^^^ 
tr—2mx-\ S-. 

Extracting  the  roots  of  the  first  and  last  members,  we  hav6 
FM=dr(a ). 


OF  CURVES.  103 


Putting  r  in  the  place  of  FM,  and  r  cos.  w-fm  in  the  place  of  x, 
and  reducing,  we  get  for  the  two  values  of  r, 


^ss/I^       r  — 

€^ — rn? 

and 

r  = 

««- 

-m^ 

the  values  of  FM"' 

a—m  COS. 
and  FM. 

> 

a-\-m 

COS. 

w' 

These  values  of  r  can  be  expressed  in  a  nnore  simple  form  by 
dividing  both  numerator  and  denominator  by  a. 

For  (27)     =  — =  (12)  ijo,  'p  denoting  the  principal  param- 

eter. 

Hence        r= ^ .     or  r  = i? . 

1 cos.  w  IH cos.  w 

a  a 

If  w=90®  these  expressions  reduce  to 

(1836)  r=±ip; 

which  is  obviously  true,  for  then  r  becomes  FG  (Fig.  2)  =  ^GH  = 
(24a)  \p. 

We  may  also  simplify  the  expressions  for  r  in  (183a)  in  another 
way,  by  introducing  a  letter  that  shall  express  the  ratio  of  the  ec 
centricity  of  the  ellipse  to  the  semi-transverse  axis. 

771 

Put  this  ratio  =  e=  (14) — . 
Hence    m  =  ae,     and  rr^  —  c^^. 

Substituting  these  values  in  the  place  of  771  and  771^  and  dividing 
both  numerator  and  denominator  by  a,  the  expressions  become 

(183c)  r  =  -- '—,     and    r=-— ^ ^, 

1 — t;  cos  w  1+ecos.  w 

forms  jften  met  with. 


104 


ANALYTICAL  GEOMETRY 


(184)  Prop.  Vlll.     Theorem. 

The  equation   of  a  parabola  referred  to   its   vertex  is  f^^px 
in  which  p  represents  the  principal  parariieter. 

Let  PAH  be  the  parabola,  A  its  ver-  (Kg.  78.) 

tex,  and  AV  and  RV  the  co-ordinates 
of  any  point  R. 

Put  AV=a:,  and  RV=y. 

By  (12)  we  have 

X  :  y  ::  y  :  p. 
Hence    y^=px, 

(184a)  SchoL  By  a  similar  process 
we  can  obtain  the  equation  of  a  parab- 
ola referred  to  the  vertex  of  any  diam- 
eter, viz.: 

y'^=p'x'; 

\n  which  p'  represents  the  parameter  of  that  diameter,  (76a). 


(186)  Prop.  IX.     Problem. 

To  find  thp  equation  of  a  tangent  to  a  parabola. 

Let  MT  be  a  line  touch- 
mg  the  parabola  at  M,  it  is 
required  to  find  its  equation. 

Since  it  passes  through  the 
point  M,  its  equation  must 
(156)  be  of  the  form 

(185a)     y'—y=a  {x'—x)  ; 

in  which  x'  and  y'  are  the  co-ordinates  of  the  point  M,  x  and  y 
those  of  any  other  point  P  in  the  line  MT,  and  a  the  tabular  tan- 
gent of  the  angrle  MTX. 

I 


OF  CURVES. 


10ft 


By  trigonometry,     ET  :  ME  ::  R  :  tan.  MTX, 
That  is  (05),  2x'    :  y'       ::  1    :  a; 

which  solved  for  a,  gives 

2x'     2x'y'      ^       ^  2x'y'     2y' 

Substituting  the  last  value  of  a  into  (185a),  we  obtain  for  the 
equation  of  the  tangent  MT, 


(185&) 


y'-y=^^(^-a:); 


or,  by  reducing,  and  substituting  ps/  in  the  place  of  y", 
(185c)  yy'  =  ^p  (x'+x). 


(186)  Prop.  X.     Problem. 
To  find  the  polar  equation  of  a  parabola. 

Let  P  be  the  pole,  and  PX'  parallel  to  (Fig.  76.) 

AX,  the  angular  axis. 

Put  the  radius  vector  PM  =  r;  the 
variable  angle  MPX'=w;  AD  and  PD, 
the  co-ordinates  of  the  pole,  =vi  and  n; 
and  AN  and  MN,  the  co-ordinates  of 
the  point  M,  =0?  and  y. 

Then,  as  in  the  circle  (176), 

x=r  GO^.(ji-\-m,    and  y=rsin.  w+n. 

Substituting  these  values  of  x  and  y 
into  the  equation  of  the  parabola  (184), 
and  transposing,  we  obtain 

7^  sin.*w  +  2  {n  sin.  w— Jp  cos.  w)  r-\-f?'^pm^(^; 

which  solved  for  r,  gives 

n  sin.  w  —  ip  cos.  w 

r= r-f^ ± 

sin/  w 


ri86fl) 


/pm -r?      /  n 
sin.^  w      V 


sin.  w  —  ^p  COS. 


H7. 


14 


/on 


ANALYTICAL  GEOMETRY. 


which  is  the  equation  required,  the  two  values  of  r  representing 
PM  and  PM'. 

(187)  Schol.  1.     If  the  pole  be  placed  at  the  vertex  A,  the  terms 
that  contain  m  and  n  disappear,  and  the  equation  reduces  to 

-  p  cos.  w 

r  =  0,     and    r—^ 

the  latter  of  which  represents  AM. 


sin.  CO 


(188)  Schol.  2.     If  the  pole  be  placed  at  the  focus  F,  w  — 0,  and 

Substituting  this  value  of  m  into  (186a),  rejecting  the  terms  that 
contain  n,  and  recollecting  that  sin.^w  +  cos.^w=l,  we  obtain 

,      COS.  w±l 

the  positive  value  being  FM,  and  the  negative  FM^'. 
If  w  =  90®  the  last  equation  reduces  to  r  =  ±  ^p. 


(169)  Prop.  XI.     Theorem. 

TTie  equation  of  an  hyperbola  referred  to  its  axes  is 
ay—b^x^-\-a^b^=0; 
in  which  a  and  b  are  the  semi-axes. 

Let  SBH  be  the  hyperbola,  A'B  (Fig.  76.) 

and  DE  its  axes,  and  AG  and  SG 
the  co-ordinates  of  any  point  S. 

Put  AB=«,  AD=6,  AG=a;,  and 
SG=y. 

By  (86) 
SG2  :  A'G.GB  ::  AD^  :  ABl 

But  A'G  =  AG  +  AB,  and  GB  =  AG-AB;    and  consequently 
A'G.GB=(AG  +  AB)  (AG-AB)='^AG3-ABa. 


A      BpG 


•Leg.  4.  10.    Euc  2.  6,  Cor. 

•  L.,  4,  10  ;  W.,  §  377. 


OF  CU RVEb 


101 


Therefore     SG«  :  AG^-AB^  ::  AD^  :  AB«. 
That  is,         y«      :  x^-^d^  ::  l^       :  a«; 

which  can  be  readily  reduced  to  either  of  the  following  forms : 

(189a)  f=-^ia^-a^,     or  ay- b^x^-\-a^lr^=0. 

(1896)  Schol.  1.  The  equation  of  the  conjugate  hyperbola  is  evi- 
dently 

Dy  merely  interchanging  the  axes  and  co-ordinates. 

(189c)  Schol.  2.  By  a  process  similar  to  that  employed  in  this 
proposition,  we  can  obtain  from  (117)  the  equation  of  an  hyperbola 
referred  to  any  two  conjugate  diameters,  viz. : 

in  which  a'  and  b'  represent  the  semi-diameters. 


(190)  Prop.  XII.     Problem. 

To  find  the  equation  of  a  tangent  to  an  hyperbola. 

By  a  process  analogous  to  that  employed  for  the  equation  of 
a  tangent    to   an  ellipse 

<180),  and  which  will  be  (Fig.  77.) 

readily  supplied  by  the 
student,  we  obtain  the 
equation 

which  may  also,  in  the 
same  manner  as  in  (180c), 
be  reduced  to  the  form 


108 


ANALYTICAL  GEOMETRY. 


(191)  Prop   XIII.     Problem. 
To  find  the  polar  equation  of  an  hyperbola 

(Fig.  78.) 


X' 


By  the  same  process  as  in  the  ellipse  (181),  we  obtain  the  equa- 
tion 

(a"  sin.2  CO— 6*  cos.*  w)  r*+2  {c^n  sin.  w—  INn  cos.  w)  r'\'cM^  6*»i'  + 

which  solved  for  r  gives 

,,^,  .  c?n  sin.  u—l^m  cos.  w 

(191a)  r= 2~^-^ j^ 2 — =t 

a''  sin.-^  u—fr  cos.'*  w 

/—aW—a^n^-{-b^m^/a^n  sin.  u^l^m  cos.  wV 
a*sin.*w— ft^cos.^w     \  a*  sin.^  w— 6*  cos.^co  /* 


(1916)  /ScAoZ.  1.     In  the  same  manner  as  in  the  ellipse  (182),  we 

find  that  if  the  pole  be  placed  in  the  centre  A,  the  expressions  be 

come 

ab 


r=s± 


■^  —  a^  sin.^  w  +  6^  cos.*  w 


which  obviously  becomes  impossible  when  a  sin.  w  >  6  cos.  w,  show- 
ing that  in  that  case  the  radius  vector  will  not  meet  the  curve.     Il 

,  ,       .     .-,  6       sin.  w      .,        .  .    \  .^ 

a  sm.  w  =  o  COS.  w;  that  is,  if  —  = =(by  trigonometry)  tan.  w, 

a      cos.  u     ^  -^      ^ 

the  denominator  becomes  o,  which  renders  the  value  of  r  infinite. 


OF  CURVES.  109 

When  the  radius  vector  is  thus  situated  it  is  called  an  asymptote  ; 

and  it  is  evident  from  the  expression  —  =  tan.  w,  that  it  lies  in  the 

direction  of  the  diagonal  of  a  parallelogram  described  on  the  two 
axes :  as  AL. 


(191c)  Sctiot.  2.  In  the  same  manner,  as  in  the  ellipse  (183),  and 
by  reference  to  (89a),  we  find  that  if  the  pole  be  placed  at  one  of 
the  foci,  the  expressions  become 


and    r  = 


a— m  cos.  CO  a  4- ''I  COS.  CO 

which  may  be  reduced  to  a  more  simple  form  by  dividing  both 
I  umerator  and  denominator  by  a. 

For        ^^-^  =  (89a)  -^ -  (12)  -ip. 

Hence    r  = — ,    and  r  = — . 

1 COS.  w  li —  COS.  w 

a  a 

We  may  also,  as  in  the  ellipse  (183c),  simplify  the  expressions  by 
introducing  a  letter  (e)  to  represent  the  ratio  of  the  eccentricity  to 
the  semi-transverse  axis,  which  will  reduce  them  to  the  form 

(I9ld)  r=^-—^ -,     and  r  =  ---^ '—. 

1— ecos.  w  1  H-e  COS.  cj 


(192)  Prop.  XIV.     Lemma. 

Every  equation  between  two  variables  of  the  form  cy^-^-da^+e^^O 
is  the  equation  of  a  conic  section} 

The  three  constants,  c,  d,  and  e,  that  enter  into  this  equation,  rep- 
resent any  known  quantities  whatever,  whether  positive  or  negative, 
and  one  of  them  must  evidently  have  the  contrary  sign  from  the 


■  See  Appendix,  Note  C. 

*  Including  in  this  term  the  circlp  and  straight  line.     See  note  on  page  9. 


110  ANALYTICAL  GEOMETRY. 

Other  two,  or  the  values  of  x  and  y  will  be  imaginary.     Hence  the 
equation  will  reduce  to  one  of  the  three  forms 

/  cy^  -\-dx^ — 6=0. 

(192a)      \cy'^  —  dx^-\-e=o. 

\  dx^^cy'^-\-e=o. 

Multiply  each  by  e  and  divide  by  cd,  and  they  become 
d^        c         cd 

\^c         d         cd 

0  g 

Putting  -7  =  a^  and  —  =  6^  and  substituting,  they  become 

(  ay+^V-  a'^^  =  0,  the  ellipse  (1786). 
(192c)      )  ay—  hV  +  a%^  =  0,  the  hyperbola  (189a). 

(  ^a;^  —  ay  +  a^i^  =  0,  the  conjugate  hyperbola  (1896). 

If  in  the  original  equation  e  =  o,  c  and  d  must  have  contrary 
signs,  and  the  equation  will  reduce  to  y=l  —  rx,  which  is  (15i) 
the  equation  of  a  straight  line  passing  through  the  origin.  Or,  if  in 
the  first  of  (192a)  c=d,  the  equation  will  reduce  to  y^-\-a^  =  '-,  the 
equation  of  a  circle  referred  to  its  centre  (173). 

The  conditions  of  the  proposition  forbid  that  either  c  or  ^f  should 
become  zero,  for  then  would  one  of  the  variables  disappear. 

(193)  Prop.  XV.     Theorem. 

Every  equation  of  the  second  degree  between  two  variables  is  the 
equation  of  a  conic  section. 

Every  such  equation  can  be  reduced,  by  transposition,  multipli- 
cation, and  division,  to  the  form 

(193a^  Ay2  +  Ba:y  +  Ca;2  +  Dy  +  Ea;-fF=0; 


OF  CURVES  ill 

in  which  the  co-efficients,  A,  B,  C,  &c.,  represent  any  known  quan- 
tities, whether  positive  or  negative.  This  equation  solved  for  y 
gives 

(193i)  .v=-^^Bz  +  D)± 

JLv(B2_4AC)a:''  +  2  (BD-2AE)  x  + (D«-4AF) ; 

in  which  we  will,  for  the  sake  of  simplicity,  put 

B2-4AC=jo,     BD-2AE=^,     and  D2-4AF=«; 

which  will  reduce  the  equation  to  the  form 
1     ,^     .  ^.        1 


(193c)         y  =  -_(Ba;  +  D)±— v'pa;2-}-2(7x+5. 

This  value  of  y  consists  of  two  parts,  and  if  the  first  part  were 
taken  by  itself,  we  should  have  the  equation  of  a  straight  line,  viz. : 

(193^  y=-3X<B^  +  ^)  =  -2l^-S^ 

R 

in  which  —  ^  corresponds  to  a  in  the  general  form  (153),  and 

D        ^ 

—  ;rr  to  6. 
2A 

The  second  part  bemg  either  positive  or  negative,  shows  tha 
each  ordinate  meets  the  curve  in  two  points,  one  as  far  above  th^ 
line  of  which  (193^?)  is  the  equation,  as  the  other  is  below  it,  and 
consequently  that  this  line  bisects  the  curve,  or  is  its  diameter. 

At  the  points  where  this  line  intersects  the  curve,  if  at  all,  the 
second  part  of  the  value  of  y  in  (193c)  must  be  zero,  and  we 
therefore  have  for  these  points  the  equation 

(193^2)  /7a:2_^2^a;  +  5=0, 

wnich  solved  for  x  e^ives 

P      P 


112 


ANALYTICAL  GEOMETRY. 


Hence  for  the  centre,  or  point  midway  between  tlie  intersections, 
we  have 


(193/) 


x^-^' 
P' 


and  substituting  this  value  of  x  into  (193rf),  we  obtain 
(193^) 


y  =  -2l(-Sj+I»- 


Equations  (193/)  and  (193^)  make  known  the  co-ordinates  of  the 
centre  of  the  curve,  if  it  have  a  centre. 

The  preceding  discussion  will  be  better  understood  by  the  aid  oi 
a  diagram. 

(Fig.  79.) 


Let  GX'  and  IKNO  be  the  straight  line  and  curve,  of  which 
(193d)  and  (193c)  are  the  equations,  drawn  according  to  (170) 

•D 

As  —  —  represents  the  tangent  of  the  angle  which  GX    makes 
with  the  axis  of  abscissas,  reckoned  in  the  usual  direction,  we  have 

{l9Sh) 


^      *       Ar^v/    .sin.  AGX' 

•— r  =  tan.  AGX'=* ttt^t- 

2A  cos.  AGX' 


Since  the  members  of  this  equation  have  opposite  signs,  it  follows 
that  the  sine  and  cosine  of  AGX'  have  the  same  sign  when  the 
signs  of  A  and  B  are  unlike,  and  contrary  signs  when  those  o. 
A  and  B  are  alike.     In  constructing  the  diagram  the  signs  of  A 


Davies'  Leg.  Trig.  Sec.  XVII. 


OF  CURVES.  113 

&nd  B  were  supposed  to  be  alike;  otherwise  the  line  GX'  would 
ha\e  been  situated  in  the  first  and  third  quadrants,  instead  of  the 
second  and  fourth,  as  it  now  is. 

It  is  also  to  be  noticed  that  the  line  AH,  represented  in  {l9Sd) 

by  the  fraction r-,  falls  below  the  axis  of  abscissas  when  the 

signs  of  A  and  D  are  alike,  but  above  it  when  they  are  unlike. 

The  values  of  y  in  (193c)  represent  any  ordinate  to  the  curve, 
as  PF,  meeting  the  curve  at  two  points,  P  and  P';  the  first  part 
representing  FS,  and  the  second  PS  or  VS. 

The  values  of  x  in  (193e)  represent  AR  and  AT,  the  abscissas 
of  the  points  I  and  N;  and  the  values  of  x  and  y  in  (193/)  and 
(193^),  the  co-ordinates  of  the  point  A',  midway  between  I  and  N. 

We  will  now,  by  a  transformation  of  co-ordinates  (162),  refer 
the  curve  to  the  lines  GA'X'  and  LA'Y'  as  new  axes,  transferring 
the  origin  of  co-ordinates  from  A  to  A'. 

The  equations  for  transformation  (162a  and  1626)  are 

(  x'  COS.  a-t-y  cos.  a'-{-m^x, 
(193/) 

(  x-  sm.  a-ry  sm.  a'-i-7t=y,- 

,n  which,  when  applied  to  the  present  case,  a  represents  the  angle 
AGX',  a'  the  angle  ALY',  both  estimated  in  the  usual  direction,  m 
and  n  the  co-ordinates  of  A',  viz. : 

m=— ^=AL,     and  w=--i-(-B-^+D)  =  A'L; 
p  2A  p 

ana  x'  and  y'  the  co-ordinates  of  any  point  in  the  curve  referred  lo 
tne  new  axes.  Since,  as  here  drawn,*  the  angle  AGX'  terminates 
m  the  fourth  quadrant,  its  sine  is  negative  and  its  cosine  positive.* 

We  have  also  cos.  a'=cos.  270°=0,   and  sin.  a'=?sin.  270°=  —l. 


'  The  general  process  is  the  same  for  any  other  position  of  the  line  GA'X 
regard  being  had  to  the  algebraic  signs. 
*  Davies'  Leg.  Trig.  Sec.  XH. 

15 


114  ANALYTICAL  GEOMETRY. 

Hence  the  equations  in  (193/)  become 

x'  COS.  a— -^=2;, 
(193m)       \ 

^x'  sin.  a_y'_±.  (_B  J+D)=v 

Substituting  these  values  of  x  and  y  into  (193c),  and  cancelling 
equal  terms,  we  obtain 

(1937i)      —x'  sin.  a-\-y'=-—-rx'  cos.  a  drr-r  ViPa;'*cos.'a— — +  s. 
^      2A  2A  *^  ^  JO 

But  (193A)  shows  that  —r  cos.  a— —sin.  a;  consequently  (1937i) 
reduces  to 


(193o)  ?/'=±— -  y^2;'2  (.Qg  2  ^_?_  +  5. 


Squaring,  clearing  of  fractions,  and  transposing,  we  obtain 

t 
P 


(193p)  4Ay^-joa;'2  cos.^  «+^-s=0 


o2 
Or,  if  we  put  4A2=c,  — p  cos.^«=^,  and  - — s=e,  the  equation 

will  read 

cy'^-\'dx'^-^e=iQ: 

which,  by  (192),  is  the  equation  of  the  conic  section. 

(193r)  Schol.  There  is  a  single  case  not  provided  for  in  the 
foregoing  demonstration.  If  B^  =  4AC,  the  value  of  p  becomes 
zero,  and  consequently  that  of  x  in  (193/)  infinite,  showing  that 
when  that  relation  exists  the  curve  cannot  have  a  centre,  and  con- 
sequently that  the  new  system  of  co-ordinates  cannot  be  referred 
to  it.     In  that  case  (193^2)  will  become 

s 

2qx-\'S=Q,     or  a:=  — —  ; 

.vhich  will  reduce  (1936?)  to  the  form 

^  B^        D 
^  ~  4A</      2A* 
The  last  two  equations  make  known  the  co-ordinates  of  the  point 
w^here  the  line  GX',  of  which  (193rf)  is  the  equation,  and  which 


OF  CURVES. 


115 


lemains  unchanged,  intersects  the  curve,  as  at  N  in  the  diagram 
We  may  therefore  refer  the  curve  to  the  lines  NX'  and  NT  as 
new  axes,  transferring  the  origin  to  N  instead  of  A'. 

The   equations    for    transformation    corresponding   to   those   in 
(193^),  will  be 


«'  cos.  a——'  =  Xy 
2q 


—a;'  sin.  a  —  y'- 


Bs        D 


■^^y- 


4Aq      2A 

By  substituting  these  values  of  x  and  y  into  (193c),  and  taking 
the  same  steps  we  did  to  obtain  (1937i)  and  (193o),  we  get 

y'=  ±  —J-  V2qx'  cos.  a  ; 

which  squared,  gives 

,2    q  cos.  a    ,       ,         ^^.        ,  c     9  ^^^-  ^ 
y'^^l—^x'=p'x,  puttmg  p'  for     ^^^   . 

Hence  the  curve  is  a  parabola  (184a). 


ri94)  Prop.  XVI.     Problem. 

The  equation  of  the  tangent,  to  any  cuTve  being  given,  to  find  tiu 
equation  of  the  normal. 

Let  CMB  be  the  curve,  MT  the  tangent,  and  MG  the  normaL 

(Fig.  80.) 


^B 


D      A 


ri6  ANALYTICAL  GEOMETRY. 

Since  Doth  these  lines  pass  through  the  point  M,  their  equation's 
will  (156)  be  of  the  form 

S  y'—y^^  {x'—x),     the  equation  of  MT, 
i  y' — y=a'  {x'—x)^     the  equation  of  MG; 

.n  which  x'  and  y'  are  the  co-ordinates  of  the  point  M,  x  and  y 
those  of  any  other  point  in  the  line  MT  or  MG,  a  the  tangent  of 
the  angle  MTX,  and  a'  the  tangent  of  the  angle  MGX.  Moreover, 
since  GM  is  perpendicular  to  MT,  we  have,  by  (161), 

(1946)  H-aa'=0. 

By  substituting  into  (1946)  the  value  of  a  taken  from  the  given 
equation  of  MT,  we  may  obtain  the  value  of  a\  and  this  value 
again  substituted  into  the  above  equation  of  MG,  gives  us  the  equa- 
tion required. 

To  illustrate,  we  will  suppose  the  curve  to  be  a  circle  referred  to 
its  centre.     The  equation  of  its  tangent  (1756)  is 

x' 

y'-y=.--{x'--x): 

x' 
m  which ;  corresponds  to  a  in  the  general  loimula  (194rt) 


Substituting  this  value  of  a  into  the  formula  1+aa  =o,  we  obtain 
which  solved  for  a'  gives 


l-^a'=0; 

y 


x' 

Substituting  this  value  of  a'  into  the  equation  of  MG  in  (194a), 
we  have 

y'^y=L{x'-x)) 

which  is  the  equation  of  the  normal. 

Example.    Find  the  equation  of  the  normal  to  the  ellipse,  parab- 
ola, and  hyperbola. 


OF  CURVES. 


Ill 


ay 
Ans.     y'—y=z  —^  (^x'—x),  for  the  ellipse. 

y'—y=' — -  {x'—x),  for  the  parabola 

c^y' 
y'—y=—rrT  (^'~^)>  ^^^  ^^e  hyperbola. 


(Iif5)  Prop.  XVII.     Problem. 

The  equation  nf  a  tangent  or  normal  to  a  curve  being  given,  tc  find 
the  points  where  it  intersects  the  axes  of  reference. 

Let  CMB  be  the  curve,  % 

AX  and  AY  the  axes,  MT 
the  tangent,  and  MG  the 
normal  at  the  given  point 
M;  it  is  required  ta  find 
the  distances  AT,  AH,  AG, 
and  AF. 

The  equation  of  the  tan- 
gent is  (156)  of  the  form 

y'-y-a{x'-x)', 

in  which  x'  and  y'  are  the  co-ordinates  of  the  given  point  M,  and  x 
and  y  those  of  any  point  P  in  the  tangent  MT.  As  this  equation  is 
true  wherever  the  point  P  be  taken,  we  may  suppose  P  to  move 
towards  T  till  the  ordinate  PD  becomes  o,  and  a:=AT.  The  equa- 
tion will  then  read 

y' —o=a  {x'  —x)  ; 

m  which  all  the  quantities  except  x  are  known,  and  consequently  its 
value  may  be  found,  which  gives  the  length  of  AT. 

In  like  manner,  by  moving  P  the  other  way  till  it  coincides  with 
H,  we  shall  have  a:=o,  and  y=AH,  which  reduces  the  equation  to 

y'—y=a  {x'-o). 

This  solved  for  y  makes  known  the  length  of  AH. 


118  ANALYTICAL  GEOMETRY. 

The  lengths  of  AG  and  AF  are  found  in  the  same  way  by  using 
the  equation  ♦of  the  normal  (194),  and  supposing  the  point  R  to 
move  first  to  G  and  then  to  F. 

Ex.  1.  Let  us  suppose  the  curve  to  be  a  parabola  whose  prin- 
cipal parameter  is  9  inches,  the  abscissa  AE  =  4  inches,  and  the 
ordinate  MN  6  inches.*  It  is  required  to  find  the  lengths  of  AT, 
AH,  AG,  and  AF. 

By  the  equation  of  the  tangent  (185),  and  normal  (194),  we  have 
For  the  point  T,        y'_o=-^  (x'-x)  ;  that  is.  6— 0=^^^  (4— x). 

Hence     a?=— 4=AT. 
For  the  point  H,        y'— 1/=-^  {x'—o) ;  that  is,  6-y=^^  (4— o). 

Hence    y=3=AH. 

2y' 
For  the  point  G,    y'-—o  = —  (x'-x) ;  that 4s,  6— o=  —  y  (4— a;; 

Hence     a; =8^= AG. 

2v' 
For  the  point  F,    y'—y= (x'—o) ;  that  is,  6-y=— ^  (4— o). 

Hence     2/=llJ=AF. 

Ex.  2.  Let  the  curve  be  a  circle  referred  to  its  centre  (173),  the 
radius  being  10  inches,  the  abscissa  —6  inches,  and  the  ordinate  8 
inches.     Required  as  above. 

Ans.     AT=-16f  inches,  AH=12i  AG=0,  and  AF=0. 

We  discover  from  this  example  that  any  normal  to  a  circle  passes 
through  the  centre. 

Ex.  3.  Let  the  curve  be  an  ellipse  referred  to  its  axes  (178), 
the  transverse  axis  being  10  inches  and  the  conjugate  8  inches. 


•  In  this  problem  and  several  which  follow,  it  would  be  sufficient  to  give  the 
value  of  but  one  of  the  co-ordinates,  x  or  y,  as  the  other  could  be  found  from  it 
by  means  of  the  equation  of  the  curve ;  but  it  would  render  the  solutions  mora 
complex  when  the  equation  is  above  the  first  degree,  since  more  than  one  value 
could  be  found  that  would  satisfy  the  conditions  of  the  problem. 


X 


OF  CURVES.  119 

the  abscissa  3  inches  and  the  ordinate  3^  inches.     Required  as 

above. 

Ans,     AT=8J  inches,  AH=5,  AG=1^,  and  AF=  — If 

(195a)  Cor.  We  are  enabled  by  this  proposition  to  determine 
the  area  of  tlie  triangle  AHT  or  AGF,  which  the  tangent  or  nor- 
mal forms  with  the  axes  of  reference. 


(196)  Prop.  XVIII.     Problem. 

ITwo  curves  ^whose  equations  are  known  A^nter sect  one  another^  to  efo- 
termine  the  point  of  intersection. 

Let  BMC  and  DME  be  the  two  curves,  (Fig.  81.) 

Y 

and  M  the  point  of  intersection. 

At  the  point  M  the  co-ordinates  AN  and 
MN  are  common  to  both  curves,  so  that 
we  have  but  two  unknown  quantities,  x  and 
y;  and  we  have  two  equations,  viz.:  the 
equations  of  the  two  curves,  by  which  to 

find  these  values.     Solve  these  equations  for  x  and  y,  and  we  have 
the  co-ordinates  of  the  point  required. 

If  the  curves  are  not  referred  to  the  same  axes,  it  will  be  neces- 
sary as  a  preliminary  step  to  reduce  them  to  the  same,  in  the  manner 
detailed  in  (162  to  168). 

Ex.  1.     Let  the  curves  be     y^=^px,     a  parabola  (184) ; 
and    x^-f  y^=^,      a  circle  (173)  ; 
in  which  ^=10,  and  r=12. 

Solving  these  equations  for  x  and  y,  we  obtain 

y=  V—ip^±p  Vr^-fi/  =  ±  Vso. 

Ex.  2.     Let  the  curves  be 

ay  -^-b^x"^  ^a^b^  =0,     an  ellipse  , 
/and     a'^y^-b'^x^-\- a%'^  =  0,     an  hyperbola ; 
in  which  «  =  10,  a' =8,  6=8,  and  6'=  6. 

Ans.     X  =  ±9,12  nearly ;  and  y  =  ±3,3  nearlj. 


(Fig.  82.) 

"i 

\^ 

^^-C 

T  ^^'^^ 

\./ 

X 

120  ANALYTICAL  GEOMETRY. 


(197)  Prop.  XIX.     Problem. 

DetermiTie  the  angle  formed  hy  the  intersection  oj  two  curves,  tht 
equations  of  whose  tangents  are  given. 

Let  BMC  and  DME  be 
the  two  curves,  and  MS  and 
MT  tangents  to  them  at  the 
point  of  intersection  M. 

The  equations  of  MS  and 
MT  make  known  the  tan- 
gents  of  the    angles   MSX 

and  MTX,  from  which  we  can  obtain,  by  (160),  the  angle  SMT^ 
which  is  the  same  as  that  made  by  the  curves  at  the  point  of  inter 
section. 

Ex.  1.     Let  DME  be  a  parabola,  and  BMC  an  hyperbola. 

Then  (190)  the  equation  of  MS  is    y'—y  =  -^-,  {x'-x) 

ay 

And  (1856)  the  equation  of  MT  is    y'-y  =  ^,  {x'—x). 

Zy 

Hence     tan.  MSX  =  ^„     and   tan.  MTX  =  -f-. 
ay'  2y' 

■  Substituting  these  values  in  the  place  of  a  and  a'  in  the  formula 
(160),  reducing,  and  for  y'^  substituting  its  equal  (184)  px',  we  get 

tan.  SMT  =  — 2—^. 
(2a^  +  o )  y 

All  the  quantities  in  this  expression  being  supposed  to  be  known,, 
we  have  the  angle  SMT. 

Ex.2  and  3.    Find  the  angles  dt  which  the  curves  in  Prop.  XVIII 
intersect. 

Ans,     The  parabola  and  circle  at  an  angle  of  71°  0'  58''. 
The  ellipse  and  hyperbola  at  an  angle  of  62°  14'  4" 


OF  CURVES. 


121 


(198)  Prop.  XX.  Problem. 

Given  the  equation  of  a  curve  and  of  its  tangent,  to  find  the  point 
on  the  curve  at  which,  if  a  tangent  he  drawn,  it  will  make  a  given 
angle  with  the  axis  of  abscissas. 

(Fig.  83.) 


Let  BMC  be  the  given  curve,  and  MTX  the  given  angle. 

Put  tan.  MTX  =  m. 

The  equation  of  MT  makes  known  the  value  of  MTX  in  terms 
of  x',  y\  and  constants,  vs^hich  being  put  equal  to  m  gives  us  one 
equation,  and  this,  together  with  the  equation  of  the  curve,  will 
enable  us  to  find  the  values  of  the  co-ordinates  x'  and  y'. 

Ex.  1.     Let  the  equation  of  the  curve  be  x'^  +  y'^  —  r^^O,  a 

x' 
circle,  the  equation  of  whose  tangent  (1756)  is  y'—y=^ -{x'—x). 

x' 
Hence     tan.  MTX= -  =  m. 

y 

Solving  this  and  the  equation  of  the  curve  for  a:'  and  y',  we  get 


mr 


^m^-^\ 


and    y'=  ± 


N^m^H-l 


Ex.  2.     Let  the  curve  be  a  parabola,  whose  parameter  is  9 
inches,  and  the  given  angle  20°. 

.4715.     a;'=  16,984,     and  y'=12  3«4. 


122  ANALYTICAL  GEOMETRY. 


(199)  Prop.  XXI.     Problem. 


To  find  where  the  tangent  to  a  curve  at  a  given  point  will  intersect 
another  curve,  the  equation  of  the  second  curve,  and  that  of  the 
tangent  to  the  first  being  given. 

This  is  merely  a  particular  case  of  Prop.  XVIII.,  and  is  solved 
in  the  same  manner,  the  two  equations  which  determine  the  values 
of  X  and  y  being  that  of  the  tangent  line  and  of  the  second  curve. 

Ex.  1.  Let  a  tangent  to  a  parabola  intersect  a  curve  whose 
equation  is  xy  =  a. 

The  equation  of  a  tangent  to  a  parabola  (1856)  is 

Solving  these  two  equations  for  x  and  y,  and  substituting  px  lu 
the  place  of  y'^,  we  obtain 

x=-'^x'±  /2ay'-)r^x'^,     and  ?/= = 

^       -^p  --\x'±  /2ay[-\-\x'^ 

If  we  give  numerical  values  to  the  letters,  as  a=10  inches,  jo=9, 
x'—4,  and  y'=6,  we  have  a:= +  2,163  inches  nearly,  or  —6,163 
nearly;  and  y=+4,62  nearly,  or  —1,62  nearly. 

Ex.  2.  Let  a  tangent  to  a  circle  whose  radius  is  10  inches  meet 
a  parabola  whose  parameter  is  9,  the  co-ordinates  of  the  point  ol 
tangency  being  x'=S,  and  y'=6,  and  the  vertex  of  the  parabola 
and  also  the  origin  of  the  co-ordinates  being  the  centre  of  the 
circle. 

Ans.     ar=6,68,  or  23,36;     and  y=7,756,  or  —14,5. 


(200)  Prop.  XXII.     Problem. 

To  determine  the  distance  from  a  given  point  to  a  tangent  to  a 
curve,  the  equation  of  the  tangent  being  given. 


OF  CURVES. 


123 


Let  P  be  the  given 
point,  and  TM  the 
given  tangent  to  the 
t>jrve  BMC  at  the 
point  M,  it  is  required 
to  find  the  distance 
PH  perpendicular  to 
TM ;  and  as  a  prelim- 
inary step  to  find  AE 
and  EH,  the  co-ordi- 
nates of  the  point  H ;  for  if  these  be  known,  the  distance  PH  can 
be  determined  by  (159). 

Produce  PH  till  it  meets  the  axis  of  abscissas  in  F. 

Put  AN  =  a:',  MN  =  y',  AE  =  a:,  EH  =  y,  AD  =  m,  DP  =  n,  taD 
HTX=a,  and  tan.  HFX=a'. 

Since  THF  is  a  right  angle  we  have,  by  (161),  the  equation 

l+fla^=0,    or  a'= ; 

a 

which  makes  known  the  value  of  a',  that  of  a  being  given  in  tne 
equation  of  TM. 

Since  the  line  TM  passes  through  the  given  point  M,  and  PF 
through  P,  their  equations  will  be 

(200a)  y' ^y—a  {x'—x),  the  equation  of  TM. 

(2006)  n  —y=a'  {m—x),  the  equation  of  PF. 

These  equations  solved  for  x  and  y  make  known  the  co-ordiaates 
of  the  point  H,  and  we  can  then  find  the  distance  PH  by  the  for- 
mula at  (159),  viz.:  PH=  V{m—xf-^ {n—yy. 


Ex.  1.     Let  the  curve  be  a  circle  refen*ed  to  its  centre,  the  equa- 
tion  of  whose  tangent  is  (1756) 

x' 
(200c)  y'-'y= -jipc  —  x),  the  equation  of  TM 

in  which ^  corresponds  tj  a  in  the  general  form 


124  ANALYTICAL  GEOMETRV. 


Substituting  this  value  of  a  into  the  formula  l+aa'=0,  and  solr- 
ing  for  a',  we  get 


x' 


This  value  of  a!  substituted  into  the  equation  of  PF  (2006),  gives 
(2006?)  n-y=^—  (m—x). 

Solving  equations  (200c  and  200c?)  for  x  and  y,  and  recollecting 

(173)  that  x''-\-y'^=?^,  we  obtain 

,  ,  (my' — nx')  y'           ,             ,.  (nx'  —  ?ny')x 
x=x  +  ^   ^    ^     '^ ,     and    y=y+^ -^Z^— 

If  r=10,  x'=S,  y'=6,  m=l2,  and  n=5,  we  shall  find 
a:  =  9,92,     y  =  3,44,     and  PH  =  2,6. 

Ex.  2.  A  comet  moving  from  C  towards  B  in  the  parabolic 
orbit  CMB,  whose  parameter  is  150  millions  of  miles,  its  vertex  at 
A,  and  its  transverse  AX,  arrives  at  the  point  M,  where  the  ordi- 
nate MN.is  100  millions  of  miles,  and  at  that  point  flies  off  from  its 
orbit  in  the  direction  of  the  tangent  MT.  The  earth,  at  the  time 
the  comet  passes  it,  is  at  P,  where  the  ordinate  PD  is  7  millions  of 
miles,  and  the  abscissa  AD  51^  millions  in  the  negative  direction.  How 
fiu-  does  the  comet  pass  from  the  earth  ?  ^^^  j^  ^^^^^^  .^ 

(201)  Promiscuous  Examples. 

1.  A  circle  whose  radius  is  10  touches  externally  an  ellipse  whose 
transverse  axis  is  10  and  its  conjugate  axis  8;  the  abscissa  of  the 
point  of  contact  referred  to  the  axes  of  the  ellipse  is  3.  Required 
the  position  of  the  centre  of  the  circle. 

Ans.  The  abscissa  of  the  centre  referred  to  the  axes  of  the  el- 
lipse is  8,145  ;  and  the  ordinate  ±11,78.* 

2.  Find  the  point  on  a  parabola,  whose  parameter  is  9,  at  which 
if  a  tangent  and  normal  be  drawn,  the  triangles  which  they  form 

•  The  negative  value  of  x  and  the  corresponding  values  of  y  are  the  co-ordinates 
of  the  centre  of  an  inscribed  circle  touching  the  ellipse  at  the  same  point. 


OF  CURVES.  125 

with  the  axes  of  reference  (195fl)  shall  be  to  each  other  in  the 

ratio  1:8.  ,  .  ^  ,  /tt"^ 

Ans.     a?=4,5,     and  y=v40,5. 

3.  Find  the  point  on  a  parabola,  whose  parameter  is  9,  at  which 
if  a  tangent  and  normal  be  drawn,  they  will  form  w^ith  the  trans- 
verse axis  a  triangle  whose  area  is  100. 

^715.     a?  =  8,92,     and  y  =  8,96. 

4.  Find  the  point  on  an  hyperbola  whose  transverse  axis  is  10, 
and  conjugate  axis  8,  at  which  if  a  tangent  and  normal  be  drawn, 
the  subtangent  will  be  to  the  subnormal  in  the  ratio  2  :  5. 

Ans.     a?  =  5,8,     and  y  =  2,35. 

5.  Any  number  of  curves  intersect ;  it  is  required  to  find  liae 
distance  between  any  two  points  of  intersection.* 

Let  there  be  four  curves  whose  equations  are  :  1st,  y=a;— 4  ;  2d, 
y^-lx\  3d,  x^-\-y^=\QO;  and  4th,  y  (j:-3)=8. 

Find  the  distance  from  the  intersection  of  the  first  and  fourth 
above  the  axis  of  abscissas,  to  the  intersection  of  the  second  and 
third  below  the  axis  of  abscissas.  Also,  find  the  distance  from  the 
intersection  of  the  first  and  second,  to  the  intersection  of  the  secont^ 
and  third,  both  above  the  axis  of  abscissas. 

Ans.     The  distances  are  9,44  and  7,31. 

6.  Find  whether  the  lines,  whose  lengths  were  found  in  the  last 
example,  intersect ;  and  if  so,  where,  and  at  what  angle.** 

Ans.  They  will  intersect,  if  produced,  at  an  angle  of  71°  50'  17" ; 
and  the  abscissa  of  the  point  of  intersection  will  be  5,91,  and  tho 
ordinate  6,55. 

*  The  coordinates  of  the  points  of  intersection  being  found  by  (196),  the  dis- 
tance may  be  found  by  (159). 
"  Each  of  the  lines  passes  through  two  given  points ;  hence  (167)  their  equations 

v"  ~~  v' 
are  of  the  form  y'—y=l-^ — i.  (x'  —  x),  in  which  x\  y',  x",  and  y"  are  the  co-ordi- 

X    —  X 

nates  found  in  the  last  example.     Substituting  the  numerical  values  in  each,  we 
shall  have  two  equations,  by  means  of  which  the  point  of  intersection  may  be  found 

by  (196),  and  tlie  angle  by  (160),  the  values  of  the  fraction  ^  ~^ .  in  the  twc 
equations,  corresponding  to  a  and  a'  in  the  formula.  x  —x 


126 


ANALYTICAL  GEOMETRY. 


CHAPTER   III. 


OF    LINES    IN    SPACE 


(202)  If  three  planes  of  indefinite  extent  intersect  each  other, 
as  in  the  figure,  they  will  divide  all  space  into  eight  parts;  and 
as  in  a  plane  the  position  of  a  point  is  determined  by  drawing 
ordinates  to  two  given  axes  lying  in  the  plane,  so  tne  position  of 
any   point   in   space   may  be 


(Fig.  86.) 


determined  by  drawing  ordi- 
nates to  these  three  co-ordi- 
nate planes,  as  they  are  usually 
styled.  For  the  sake  of  sim- 
plicity, the  three  planes  are 
usually  taken  at  right  angles 
to  each  other  when  practic- 
able, as  represented  in  the 
figure,  and  in  our  future  dis- 
cussions they  are  to  be  so 
regarded.  The  plane  STVH 
is  called  the  horizontal  plane, 
and  the  other  two  the  verti- 
cal planes.  The  lines  X'AX, 
Y'AY,  and   Z'AZ,  in   which 

the  planes  intersect,  are  called  axes ;  and  the  point  A,  in  which  the 
axes  intersect,  the  origin.  Ordinates  in  the  several  directions,  AX, 
AY,  and  AZ,  are  denoted  respectively  by  the  letters  x.  y,  and  z;  and 
those  in  the  opposite  directions  by  the  same  letters  wiin  the  negative 
sign  prefixed,  in  the  same  manner  as  in  equations  of  lin6s  in  a  plane. 
Unlike  those  equations,  however,  AZ  is  usually  taken  as  the  axis  oi 
abscissas,  and  the  other  two  as  axes  of  ordinates. 


OF  LINES  IN  SPACE.  127 

(203)  Instead  of  giving  the  lengths  of  the  ordinates  themselves 
to  determine  the  position  of  a  point,  we  may  give  the  measure  ot 
ihem  on  the  axes  to  which  they  are  parallel,  in  the  same  manner  as 
in  the  plane.  Thus  the  position  of  P  may  be  determined  by  giving 
the  lengths  of  PD,  PD',  and  PD",  or  by  giving  the  lengths  of  AB. 
AC,  and  AE. 

(204)  The  points  where  the  ordinates  of  a  point  in  space  meet 
the  several  co-ordinate  planes  are  called  projections ;  and,  in  like 
manner,  the  lines  traced  upon  the  co-ordinate  planes  by  an  indefinite 
number  of  ordinates  let  fall  upon  them  from  any  line  in  space 
whether  straight  or  curved,  are  called  the  projections  of  that  line. 
Thus  D  is  the  projection  of  the  point  P  upon  the  plane  AZFG, 
D'  its  projection  upon  ANMZ,  and  D"  its  projection  upon  AIHL. 
Also,  the  lines  ED'  and  CD''  are  the  projections  of  the  line  PD  upon 
the  two  latter  planes.  Being  perpendicular  to  the  plane  AZFG,  it 
is  projected  upon  it  in  a  point  at  D ;  but  if  it  were  oblique  it  would 
be  projected  into  a  line  upon  this  plane  also.  The  plane  in  which 
both  a  line  and  its  projection  lie  is  called  the  projecting  plane. 

(205)  Def,  A  cylindrical  surface  is  one  generated  by  a  straight 
line  moving  parallel  to  itself,  while  its  extremity  describes  a  curve. 

(206)  Def.  A  conical  surface  is  one  generated  by  a  straight  Ime 
passing  through  a  fixed  point,  while  its  extremity  describes  a  curve. 
ft  obviously  consists  of  two  parts  united  only  at  the  fixed  point  or 
vertex.     The  two  parts  are  called  sheets  or  nappes. 

(207)  The  moving  line  in  (205)  and  (206)  is  called  the  genera- 
trix, and  the  curve  the  directrix. 

(208)  Def.     A  conoid  is  a  solid  generated  by  the  revolution  of 
either  of  the  conic  sections  about  one  of  its  axes,  and  is  either  an 
ellipsoid,  a  paraboloid,  or  an  hyperholoid,  according  as  the  genera 
trix  is  an  ellipse,  a  parabola,  or  an  hyperbola. 


128 


ANALYTICAL  GEOMETRY. 


F 

(Fig.  87.) 

z                    » 

ji^ -^. 

A 

lliiil      c*.v% 

(209)  Prop.  I.     Theorem. 

The  projections  of  a  point  or  straight  line,  or  either  two  of  the  t<h 
ordinate  planes,  determine  its  position.*^ 

Firstly.  Let  C  be  the  point, 
and  c  and  c'  its  projections 
upon  the  vertical  co-ordinate 
planes  AZFY  and  ANMZ. 

At  the  points  c  and  c'  draw 
perpendicular  to  their  respec- 
tive planes  the  lines  cC  and  c'C 
Since  the  point  C  must  by  (204) 
lie  in  both  these  perpendiculars, 
their  mutual  intersection  must 
determine  its  position. 

Secondly.  Let  CD  be  the  straight  line,  and  cd  and  c'd'  its  pro- 
jections upon  the  vertical  planes. 

Through  cd  and  c'd'  draw  the  planes  cCDd  and  c'CDd'  perpendic- 
ular to  the  co-ordinate  planes,  and  since  the  line  CD  must  lie  in  both 
these  planes,  their  mutual  intersection  must  determine  its  position. 

(209a)  Schol.  This  proposition  applies  also  to  any  plane  curve ; 
for  such  a  curve  must  lie  in  each  of  two  right  cylindrical  surfaces, 
the  projections  upon  the  co-ordinate  planes  being  the  directrices. 


S''^ 


(210)  Prop.  IT.     Theorem. 

The  equations  of  a  straight  line  in  space  are 
x  =  az-{'a,  and  y  =  hz-]r^'t 
in  which  x,  y,  and  z  represent  the  co-ordinates  of  any  point  m 
the  line,  a  and  h  the  tangents  of  the  angles  which  the  projections 
of  the  line  upon  the  vertical  planes  make  with  the  axis  of  ab- 
scissas, and  a  and  fi  the  parts  of  the  axes  of  ordinates  intercepted 
between  these  projections  and  the  origin. 

'  If  the  line  is  parallel  to  one  of  the  co-ordinate  planes,  one  of  the  projectiona 
nust  be  taken  on  that  plane. 


OF  LINES  IN  SPACE.  129 

Let  CD  (Fig.  87)  be  the  straight  line,  and  cd  and  dd'  its  projec- 
tions. 

Also  let  P  be  any  point  in  CD,  and  m  and  n  its  projections 

Produce  c'd'  and  cd,  if  necessary,  till  they  meet  the  axis  of  ab- 
scissas in  T  and  S,  and  draw  the  ordinates  ms  and  ns. 

Then  ms=Vn  =  x,  ns=Vm  =  y,  and  by  (203)  A5  =  z. 

Put  tan.  c'TZ  (the  angle  being  estimated  from  Z  towards  the 
right)  =  a. 

Put  tan.  cSZ  (the  angle  being  estimated  from  Z  towards  the  left) 
=6. 

Also  put  AH=a,  and  AG=/S. 

Then,  since  Z'AZ  is  the  axis  of  abscissas,  AX  the  axis  of  ordi- 
nates, and  X  and  z  co-ordinates  of  a  point  m  in  the  line  c'd',  situated 
in  the  plane  ZN,  the  equation  of  c'd'  is  by  (153) 

x  =  az-\-a. 
In  like  manner  the  equation  cd  is 

y  =  bz+l3. 

These  equations  determine  the  positions  of  two  projections  of  CD, 
and  hence  (209)  that  of  CD  itself 

(211)  SchoL  1.  Of  the  four  constants,  a,  a,  b,  and  ft  that  enter 
into  these  equations,  it  is  obvious  that  if  none  were  known,  nothing 
could  be  determined  in  regard  to  the  position  of  the  line  to  which 
the  equations  referred.  If  a  only  were  known,  it  would  fix  the  in- 
clination of  the  plane  cCT>d'  to  the  co-ordinate  plane  AF,  but  not 
its  position,  since  an  indefinite  number  of  planes  might  be  drawn 
parallel  to  it.  If  a  and  a  only  were  known,  the  precise  position  of 
the  plane  c'CBd'  could  be  determined,  but  nothing  in  regard  to  the 
position  of  the  line  CD  in  the  plane.  If  a,  a,  and  b  were  known, 
they  would  fix  the  direction  of  the  line  CD  in  the  plane ;  but  still 
there  might  be  an  indefinite  number  of  lines  drawn  parallel  to  it  in 
the  plane,  which  would  satisfy  the  equations  equally  well.  But 
lastly,  if  a,  a,  b,  and  (3  are  all  known,  they  Hmit  the  Hne  to  a  single 
position,  as  already  shown. 

17 


130  ANALYTICAL  GEOMETRY. 

(211a)  Schol.  2.  Since  a  determines  the  inclination  of  the  plane 
c'CDd\  and  b  the  direction  of  the  Hne  CD  in  that  plane,  the  two 
together  must  determine  the  direction  of  any  line  in  space ;  so 
that  if  these  letters  have  the  same  value  in  the  equation  of  any  one 
line  that  they  have  in  any  other,  the  lines  which  the  equations  rep- 
resent must  be  parallel,  whatever  may  be  the  values  of  a  and  ^. 

(212)  Prop.  III.     Theorem. 

TTie  equations  of  a  straight  line  passing  through  a  given  point  in 

space  are 

x'—x^a  {z'^z),     and  y'—y  =  b  (z'—z); 

in  which  a?',  y',  and  z'  denote  the  co-ordinates  of  the  given  point, 
and  the  other  letters  as  in  the  preceding  proposition. 

The  same  construction  remaining  as  in  the  last  proposition,  let  C 
be  the  given  point,  and  c'  and  c  its  projections.  Draw  the  ordinates 
c  t  and  ct. 

Then     c't  =  Cc  =  x',     ct  =  Cc'  =  y\     and  by  (203)  A^  =  z'. 

Hence,  in  the  same  manner  as  in  (156),  the  equations  of  the  pro- 
jections c'd'  and  cd  are 

a;'— a;=a  (2'— z),     and  y'^y=h  (z'  — z)  ; 

which  determine  the  position  of  the  line  m  the  same  manner  as  in 
the  last  proposition. 

(212a)  A  straight  line  passes  through  a  point  in  space  whose  co- 
ordinates are 

a;'=7,     t/'=8,     and  2'=9; 

its  projection  upon  the  plane  ZN  crosses  the  axis  of  abscissas  at  an 
angle  of  58°,  and  its  projection  upon  ZY  at  an  angle  of  45°.  Re- 
quired the  area  of  the  triangles  which  the  projections  form  with  the 

co-ordinate  axes  (195a). 

Ans.     17,1 125  and  ,5. 


OF  LINES  IN  SPACE.  131 


(213)  Prop.  IV.     Theorem. 

IVis  equations  of  a  straight  line  passing  through  two  given  points 

m  space  are 

^1 x'  v" V' 

x'-x=— j{z'-%\     and   y'-y=^       ^   {^'—^)\ 

in  which  x",  y",  and  z"  are  the  co-ordinates  of  one  point,  and 
x\  y ,  and  z'  of  the  other. 

In  the  same  manner  as  in  the  two  preceding  propositions,  it  may 
be  easily  shown  that  the  projections  of  the  line  upon  the  vertical 
planes,  are  straight  lines  passing  through  two  given  points  in  chose 
planes,  the  co-ordinates  of  the  two  points  being  for  the  one  projec- 
tio;i  (x",  z")  and  (af^  z'),  and  for  the  other  projection  (y",  z")  and 
(y',  z').     Hence,  by  (157),  the  equation  of  one  projection  is 

x"  —  x' 

i 
y"—  y' 
and  for  the  other  y>—y—^—^ — ^  (2^'— 2;). 

which  together  determine  the  position  of  the  line  in  question. 

(213a)  Ex.  A  straight  line  passes  through  two  points  in  space 
whose  co-ordinates  are  j?'=5,  y'  —  fi,  z'=8,  and  a:"=10,  y=4,  and 
z"=6.  Required  the  points  where  its  projections  on  the  vertical 
planes  cross  the  axis  of  abscissas,  and  at  what  angle. 

Ans.  One  projection  crosses  above  the  origin  at  a  distance  of 
10,  and  an  angle  of  116°  34';  and  the  other  below  the  origin  at  a 
distance  of  2,  and  an  angle  of  26°  34'. 


(214)  Prop.  V.     Theorem. 
The  distance  between  two  points  in  space  is 


V{x'-xy-\-{y'-yy-\-(z'-zy; 

in  which  a:',  y',  and  z'  are  the  co-ordinates  of  one  point,  and  at, 
y,  and  z  of  the  other. 


132 


ANALYTICAL  GEOMETRY. 


If  through  each  of  the  points  three  planes  be  made  to  pass,  par- 
allel to  the  co-ordinate  planes,  it  is  obvious  that  they  will  by  their 
mutual  intersection  form  a  parallelopiped,  of  which  the  distance  be- 
tween the  two  points  will  be  the  diagonal,  and  whose  edges  will  be 
the  difference  of  the  corresponding  ordinates,  viz. :  (x'—u-),  (y'—y), 
and  (z'—z).  But  the  square  of  the  diagonal  of  a  parallelopiped  is 
equal  to  the  sum  of  the  square  of  its  edges,.*  Hence,  if  we  let  D 
represent  the  distance  between  the  points,  we  shall  have 

(214a)  I)^={x'--xy+(t/'-yy-{-{z'-zy ; 

and  extracting  the  root, 

D=  ^{x'-xy^{y'-yy+(z'—zY.^ 

(2146)  Ex.  Required  the  distance  between  two  points  in  space; 
the  co-ordinates  of  one  being  8,  9,  and  10,  and  of  the  other  3,  4, 
and  5.  Ans.     8,66. 

(215)   Prop.  VI.     Theorem. 

The  tangent  of  the  angle  included   between  two  straight  lines  in 
space  is 

V  (a'  -  af^^fllby^i^'^  a'bf . 
\-\-aa'^-hh' 
in  which  a  and  h  represent  the  tangents  of  the  angles  that  the 
projections  of  one  of  the  lines  make  with  the  axis  of  abscissas 
and  a'  and  h'  those  of  the  other. 

Let  the  equations  of  the  two  lines  be 

i  x=^az  +a,     and  y=hz  H-/3,     for  the  one; 
(  x=a'z-{-oL',    and  y=h'z-\-^\    for  the  other. 


'  Let  AG  be  a  parallelopiped,  and  DF  its  diagonal. 

Because  DAB  is  a  right  angle  DB^^DA'+ABS  and  be- 
cause DBF  is  a  right  angle  DF^=:DB''+BF'»;  therefore  DP 
=DA^+AB''+BF«. 

"  This  proposition  may  be  illustrated  by  taking  two  points 
on  the  surface  of  an  apple,  and  while  the  apple  remains  fixed 
in  position,  cutting  it  through  each  of  the  points  in  three  di- 
rections parallel  to  three  co-ordinate  planes. 


(Fig.  88.) 


II 

G 

\ 

E 

\ 

D 

■ 

i 

^ 

OF  LINES  IN  SPACE. 


133 


(215a) 


Then  (21  la)  will  the  equations  of  two  other  lines,  as  AM  and 
.\N,  drawn  parallel  to  them  through  the  origin,  be 
x—az,     and  y=hz,     for  the  one,  AN ; 
x=a'z,    and  y=b'z,    for  the  other,  AM. 

Now  if  from  any  point  two  lines  be  drawn  parallel  to  any  other 
two  lines  in  space,  the  angle  which  the  two  latter  make  with  each 
other  is  considered  the  same  as  that  made  by  the  two  former,  even 
though  the  latter  do  not  lie  in  the  same  plane,  so  as  to  actually  in- 
tersect.* Consequently,  we  have 
only  to  determine  the  angle  formed 
by  the  two  latter  lines. 

From  any  point  N  in  AN  draw 
NM  perpendicular  to  AM,  and 
denote  the  co-ordinates  of  M  by 
jc',  y',  and  z',  and  those  of  N  by 
x",  y'\  and  z" .  And  since  the  co- 
ordinates of  A  are  zero,  we  have 
by  (214)  the  distances  AM,  AN, 
and  MN,  as  follows  : 


(Fig.  89.) 


AM=Vx'2-fy2^z'2; 


(215&) 


AN  =  v/a:"--i-y"^+2"^ 


l^MN=v^(a?''-a:')^+(y"-?/')"^+,(2"-2'A 

The  angle  AMN  being  a  right  angle,  we  have 
(215c)  AN2=AM2  4-MN2; 

ivhich,  by  transposing  AM^  and  dividing  by  it,  becomes 

(215^) 


AN^_   ^MN2 
AM^     ^~  kW 


Also,  by  trigonometry,  we  have 

tan.  MAN= 


MN 
AM' 


MN2  AN* 

(215e)      Or,   tan.»MAN=^^,=  (by  215i)  ^-1 


Legr.  6.  6,  Schol. 


134  ANALYTICAL  GEOMETRY. 

By  substituting  the  values  of  AM,  AN,  and  MN  from  (2156)  into 
(215c)  and  (215e),  and  reducing  the  former,  we  obtain 

(215/)  x'^'{'y'^+z'^=x"x'-\-y"y'+z"z'. 

x'^-\-y'^^' 


(2l5g)  tan.^  MAN=:^^^;,-Z^- 1. 


As  the  equations  (215a)  are  applicable  to  any  points  in  the  lines 
AM  and  AN,  they  will  apply  to  the  points  M  and  N,  and  for  these 
points  they  become 

x'  =az',     and  y'  =bz',     for  AM. 

for  AN. 


(  x'  =az'      and  y'  =bz' 
^  (  x"=a'z",  and  y"^h'z'\ 


These  values  of  x',  y',  x",  and  y"  being  substituted  into  (215f) 
and  (215^),  we  have 

z" 
If  now  we  substitute  the  value  of  —  from  (215^)  into  (215Q,  we 

have,  after  cancelling  the  factor  (a^+^^+l), 

(315m)     tan.  MAJM (i  +  aa'+bb'Y 

which  can  be  reduced  to  the  form 

tan.^MAN=  (-''--')'+ (^'-^)'+(f'-'»'^)'; 
{l-\'aa'-^bb'y 


/oirc  \     n       .       TVTATVT      ,^(a'-af-\-(b'-bf-]-{a'b-ab'f^ 

(21571)     Or,     tan.  MAN=±— ^^ ^-4-t ,  .  ,' , ^;* 

l+aa'+bb' 


•  The  formula  given  by  other  authors  is 

l+aa'+bb' 


COS.  MAN=- 


V(a'«+6''+l)  (a«+6'+l)' 

but  we  have  preferred  the  one  we  have  adopted  because  it  is  equally  simpie,  anu 
preserves  the  analogy  between  straight  lines  in  space  and  straight  lines  in  a  plane, 
as  will  be  seen  by  comparing  it  with  the  formula  given  in  (160). 


OF  LINES  IN  SPACE.  135 

the  two  values  being  the  tangents  of  the  two  adjacent  angles  that 
the  lines  form  with  each  other. 

(215p)  Ex.  Required  the  angle  included  between  two  lines  in 
space  whose  equations  are 

x=52+7,     and  y=3z+S,     of  the  first; 
07=42  +  10,  and  y =22  +  11,  of  the  other. 

Ans.     S**  ir. 

^216)  SchoL  1.  As  in  (101),  the  numerator  of  the  last  fraction 
must  become  zero  when  the  lines  are  parallel,  and  the  denominator 
zero  when  they  are  perpendicular  to  each  other.  But  the  numer- 
ator under  the  radical  sign  consists  of  three  perfect  squares,  each 
of  which  must  therefore  be  positive;  so  that  the  numerator  can 
become  zero  only  by  each  of  the  three  terms  of  which  it  is  com- 
posed becoming  so.  Hence  the  conditions  of  paralleUsm  between 
two  lines  in  space  are  a'=a  and  b'=b,  (as  already  shown  in  another 
way  in  (211a),)  and  of  perpendicularity 

l+aa'-[-bb'-0. 

(217)  SchoL  2.  It  can  be  demonstrated,  that  if  we  represent  the 
angles  which  one  of  the  hues  form  with  the  co-ordinate  axes  by  X, 
Y,  and  Z,  and  those  of  the  other  by  X',  Y',  and  Z',  we  shall  have 

COS.  MAN = cos.  X  COS.  X'+cos.  Y  cos.  Y'+cos.  Z  cos.  Z'.* 


(218)  Prop.  VII.     Problem. 

Two  lines  in  space,  straight  or  curved,  intersect:  determine  the  point 

of  intersection. 

At  the  point  of  intersection  the  co-ordinates  are  common  to  both 
lines,  and  may  be  found  by  solving  any  three  of  the  equations  of 
the  lines  for  x,  y,  and  z. 

'  Davies'  Analytical  Greometry. 


136  ANALYTICAL  GEOMETRY. 

Ex.     Let  a  straight  line,  whose  equations  are 

x=az4-a,     and  y=-hz-\-^, 

intersect  a  curved  line,  the  equations  of  whose  projections  are 

oi?=p  («  +  z),     and     y=b'z-]-(3', 

which  designate  a  parabola  whose  plane  is  perpendicular  to  the 
co-ordinate  plane  AYPZ  (Fig.  89). 

If  we  solve  the  first,  second,  and  fourth  equations  for  x,  y,  and  z, 
we  shall  obtain  for  the  co-ordinates  of  the  point  of  intersection, 

(3'-(3  (3'b-f3b'  .         p'-p 

If  a  different  selection  had  been  made  from  the  four  equations 
the  form  of  the  answers  would  have  been  different,  but  the  real 
values  the  same,  as  might  be  made  to  appear  by  eliminating  x,  y, 
and  z  from  the  four  equations,  and  thus  obtaining  an  equation  be- 
tween ihe  constants. 


OF  PLANE  SURFACES. 


187 


CHAPTER  IV. 


OP    PLANE    SURFACES. 


Def.     The  traces  of  a  plane  are  the  lines  in  which  it  cuts  the 
co-ordinate  planes. 


(219)  Prop.  I. 


Theorem. 

y 


The  equation  of  a  plane  is  2=—  +  -^  +  c, 

in  which  a  and  h  represent  the  tangents  of  the  angles  that  the 
traces  on  the  vertical  planes  make  with  the  axis  of  abscissas,  c  the 
part  of  the  axis  of  abscissas  intercepted  between  the  plane  and  the 
origin,  and  a;,  y,  and  z  the  co-ordinates  of  any  point  in  the  plane. 

Let  FGH  be  the  plane,  (Fig.  90.) 

P  any  point  in  it,  and  FH 
and  FG  its  traces  on  the 
vertical  planes. 

Through  P  draw  the 
plane  PMND  parallel  to 
ZX,  meeting  the  trace  FH 
in  M,  and  the  axis  AY  in 
N.  Then  will  the  lines 
ND,  AN,  and  PD  be  equal 
to  the  co-ordinates  of  the 
point  P;  that  is,  ND=a;, 
AN=y,  and  PD=2. 

From  F  draw  FL  par- 
*llel  to  AY,  and  from  L  and  M  draw  LR  and  MK  parallel  to  AX 

18 


138  ANALYTICAL  GEOMETRY. 

Then  we  shall  have 

MK=LR=ND=a;,     FL=AN=y,     and  RD=LN=AF-c. 

Produce  GF  and  HF  till  they  meet  the  horizontal  axes  in  T  and 
S.  Then,  since  the  plane  FGH  cuts  the  two  parallel  planes  MD 
and  ZX,  the  lines  of  intersection  MP  and  FG  are  parallel  ;*  and 
consequently  the  angle  PMK=GTX=the  complement  of  GFZ 
Also  the  angle  HFL=HSY=the  complement  of  HFZ. 

Put  tan.  GFZ=a,  and  tan.  HFZ=6. 

Then     tan.  PMK=cot.  GFZ=(by  trigonometry)  — , 

1  ^ 

and     tan.  HFL=cot.  HFZ=-f. 

0 

The  ordinate  PD,  or  z,  consists  of  three  parts  PK,  KR,  and  RD. 
Hence  we  have  the  equation 

(219a)  z=PK+KR+RD. 

But      PK=tan.  PMK.MK=ia;=-. 

a        a 

And    KR=ML=tan.  HFL.FL=|y=|-. 

h^      b 

And    RD=c. 
Substituting  these  values  into  (219a),  we  have 

a      0 

(219&)  Cor.  If  the  point  be  taken  in  the  trace  HF,  the  value 
of  X  becomes  zero ;  and  if  in  GF,  the  value  of  y  becomes  zero 
Hence  the  equation  of  the  trace  HF  is 

X 

and  of  the  trace  GF,  z  =  — \-c. 

a 

The  same  thing  may  be  shown  also  geometrically;  for  any  or- 
nate Mr 
trace  GF. 


dinate  MN  =  ML  +  LN=-y-+c;   and  in  the  same  manner  for  the 

0 


•  Leg  6.  10.    Euc.  Sup.  2. 14. 


OF  PLANE  SURFACES.  139 


(220)  Prop.  II.     Theorem. 

Every  equation  of  the  first  degree  between  three  variables  is  th6 
equation  of  a  plane. 

For,  by  the  ordinary  operations  of  algebra,  every  such  equation 
can  be  reduced  to  the  form 

(220a)  Aa?+B3/+Cz+D=0 ; 

in  which  A,  B,  C,  and  D  represent  any  known  quantities  whatever 
whether  positive  or  negative.     But  the  above  equation  reduces  to 

(^206)  z=__ar--y-^; 

which  corresponds  to  the  form  given  in  (219). 

If  now  we  measure  off  from  the  origin  on  the  vertical  axis  a  part 
equal  to  —j^,  and  through  the  point  thus  found  draw  in  the  ver- 
tical  planes  two  lines,  so  that  the  co-tangents  of  the  angles  which 

they  make  with  the  axis  of  abscissas  shall  be  equal  to  —  ^r,  and 

B  , 

~7Tj  the  plane  of  which  these  lines  are  the  traces  must  have  for  its 

equation  (220a)  or  (2206). 

(220c)  Schol.  It  may  be  shown,  in  the  same  manner  as  in 
(2196),  that  if  the  equation  of  a  plane  be  given  in  the  form  (220a) 
the  equations  of  its  traces  are 

Aa;+Cz+D=0,     and  By+Cz+D=0. 


(221)  Prop.  III.     Problem. 

To  find  the  equation  of  a  plane  that  shall  pass  through  three  given 

points. 

Let  the  co-oidinates  of  the  given  points  be  {x',  y',  z'),  {x"^  y",  z"), 
{x"',  y"\  z"').     The  equation  of  the  plane  must  (219)  be  of  the  form 

(S»la)  '"^f  "^f  "*'^- 


140  ANALYTICAL  GEOMETR\. 

As  the  equation  must  be  true  for  every  point  in  the  plane,  it  mu^t 
be  true  for  the  three  given  points.  We  shall  therefore  have  the 
three  following  equations,  viz. : 

a  0 

x"  v" 
a        0 

^m  ,,,11) 

^"=  — +  V  +  C. 
a         b 

By  means  of  these  three  equations  w^e  may  obtain  the  values  of 

the  unknov^rn  quantities  a,  h,  and  c,  which  substituted  into  (221a) 

will  give  us  the  equation  required.     It  can  then  be  reduced  to  the 

following  more  simple  form,  viz. : 

{x'' —x){z'  -z)--{x'  ■--x){z''  -z)_{x'''  -x){z'  -z)--(x'  -x){z'''--'t) 
(x"-x'){y'-y)-ix'-x)iy"-y')~'(x"'-x'){y'-y)-{x'-x){y''^^)' 


(222)  Prop.  IV.     Problem. 

To  find  the  equations  of  a  straight  line  that  shall  be  perpendicular 

to  a  given  plane. 

Let  the  equation  of  the  given  plane  be 

Ax-\-By+Cz-{-J)=0. 

The  plane  that  projects  the  line  upon  either  of  the  co-ordinate 
planes,  must  be  perpendicular  both  (204)  to  that  co-ordinate  plane 
and  *to  the  given  plane.  Hence,''  the  projection  of  the  line  must  be 
perpendicular  to  the  trace  of  the  plane. 

Let  the  equations  of  the  proposed  line  be 

(222a)  a:=az  +  a, 

(2226)  y=bz-\-^; 

in  which  a,  b,  a,  ^  are  evidently  unknown  quantities,  and  that  of 

the  plane 

Aa;  +  B?/  +  Cz  +  D=0. 

■  Leg.  6  16.    Euc.  Sup.  2.  17.  *  Leg.  6.  18.    Euc.  Sup.  2.  18. 


OF  PLANE  SURFACES.  141 

By  (220c)  the  equations  of  the  traces  of  the  plane  are 

C       D 
(222c)  Ax+Cz+D=0,     or  ^=— T^-~T- 

C       D 

{222d)  Bz/+Cz+D=0,     or  t/=_-y— -. 

But  the  lines  of  which  (222a)  and  (222c)  are  the  equations  lie  in 
the  same  plane,  and  are  perpendicular  to  each  otner.  Hence  we 
have,  by  (161), 

C  A 

(222e)  l  +  a(-— )=0,     ora=^: 

and,  in  like  manner,  we  have  from  (2226)  and  {222d), 
(222/)  1  +  6  (- J)=0,     or  o=g. 

These  values  of  a  and  6,  substituted  into  (222a)  and  (22^6),  give 
the  equations  required,  viz, : 

A  B 

(222^)  x=-^z-{-oi,     and  y=7T2+/3. 


(223)  Schol  The  values  of  a  and  (3  are  left  undetermined, 
which  is  as  it  should  be,  since  the  number  of  lines  that  can  be 
drawn  perpendicular  to  a  plane  is  unlimited. 


(224)  Prop.  V.     Problem. 

To  find  the  inclination  of  a  given  line  to  a  given  plane. 

Let  MN  be  the  given  line,  having  for 
its  equations 

x=az-{-a,     and  y=bz=(3; 

and  let  ABHC  be  the  given  plane,  havmg 
for  its  equation 

Ax  +  By+Cz-\-J)=0. 

From  any  point  P  in  MN  draw  PD  perpendicular  to  the  plane 
and  join  DN 


I  i2  ANALYTICAL  GEOMETRY 

The  equations  of  PD  are,  by  (222^), 

x=^z-\-cc,     and  y=—z-\-^. 

The  inclination  PND  is  the  complement  of  the  angle  NPD. 

Hence,  by  trigonometry,     tan.  PND= ^rFFTFT- 

•^       °  "^  tan.  NPD 

Now  the  equations  of  MN  and  PD  being  known,  we  may  obtain 

A 

an  expression  for  tan.  NPD  by  (215),  viz.:  by  substituting  ^  in  the 

B 

place  of  a',  and  j^  in  the  place  of  b' ;   and  then,  by  inverting  the 

fraction,  we  have  its  reciprocal,  which  is  the  tangent  of  the  angle 
required.  The  following  is  the  result,  after  multiplying  both  nu- 
merator and  denominator  by  C, 

ta„.PND=±.  Aa  +  Bb  +  C 


Cy(^-a)^  +  (g-5)^+(A^/ 


(225)  Schol.  The  numerator  of  this  fraction  must  become  zero 
when  the  line  is  parallel  to  the  plane,  and  the  denominator  zero 
when  it  is  perpendicular.  But  the  latter  can  happen,  as  was  shown 
in  (216),  only  when  the  separate  terms  become  so.  Hence,  the 
conditions  of  parallelism  between  a  line  and  plane,  whose  equation 
is  of  the  form  Aa:+By+Cz+D=0,  is 

Aa  +  B6  +  C  =  0; 
and  of  perpendicularity, 

Q=cL,     and  7T=t'; 

the  latter  being  the  same  result  that  was  obtained*  in  Prop.  IV 


(226)  Prop.  VI.     Problem. 
To  determine  the  inclination  of  two  given  plancM 
Let  the  equations  of  the  two  planes  be 


A.r+B?/+Cz+D=0, 
A'a:+B'z/+C'z+D'=0. 


OF  PLANE  SURFACES.  143 

From  any  point  in  space  let  fall  a  perpendicular  upon  each  plane 
Then,  by  (222^),  the  equations  of  the  perpendiculars  will  be 

a:=-p2+a,     and  y=p-2^4-/3;     the  equations  of  the  first 

A'  B' 

x=T^2+a',    ana  ^=^72  4-^';    the  equations  of  the  second 

The  angle  included  between  these  perpendiculars,  which  is  the 
supplement  of  the  inclination  of  the  planes,  may  be  found  by  sub- 
stituting p,  p7,  pj,  and  pj  in  the  place  of  a,  h,  a',  ana  b'  in  (215). 

Making  the  substitution  and  multiplying  both  numerator  and  de- 
nominator by  C'C,  we  ol^^ain  for  the  tangent  of  the  inclination,  or 
of  its  supplement, 

±  V  ( A^C  -  ACO^+  (B^C  -  BCQ^  +  ( A^B  -  ABQ* 
A'A  +  B'B+C'C 

(226a)  Ex.  Determine  the  inclination  of  two  planes  whose 
equations  are 

2a;  +  32/+4z— 10=0; 
Sx  —  5y—2z-\-  5=0. 

Ans.     59^  12'. 

(227)  Schol.  In  the  same  manner  as  in  the  last  proposition,  we 
see  that  the  conditions  of  parallelism  between  two  planes  are, 

a;__  A  5!__^ 

C'""C'  C'~C' 

and  of  perpendicularity, 

A'A+B'B  +  C'C=0. 


(228)  Prop.  VII.     Problem. 

To  determine  the  position  of  the  foot  of  a  perpendicular  let  fall 
from  a  given  point  in  space  upon  a  given  plane. 

Let  the  equation  of  the  plane  be 

(228a)  Aa:  +  By  +  Cz  +  D  =  0 


144  ANALYTICAL  GEOMETRY. 

Since  the  perpendicular  passes  through  a  given  point,  its  equa 
tions  must  (212)  be  of  the  form 

(2286)         x'—x=a{z'-z),     ar  1  3/'— y=6  (z'— 2)  ; 

and  since  it  is  perpendicular  to  the  given  plane,  we  have,  by  (222e) 
and  (222/), 

-g-  =  rt,     and  -Q=f>f 

which  values  of  a  and  b  being  substituted  into  (2286),  we  have 

A 
(228c)  x'—x=-^{z'-z); 

{22Sd)  y/_y=^(z'-z). 

The  three  equations  (228a),  (228c),  and  (228^0,  solved  for  x,  y. 
and  z,  make  known  the  point  required. 

(228e)  Ex.  Find  the  position  of  the  foot  of  a  perpendicular  let 
fall  from  a  point  in  space  whose  co-ordinates  are  x'=r5^  y'=Q,  ai^d 
t'=7,  upon  a  plane  whose  equation  is  4a;  +  3y  +  2z+l=0. 

/  x=—2^Q. 
Ans.     }  y=^. 
(  z=3ja. 

(229)  SchoL  After  determining  the  position  of  the  foot  of  the 
perpendicular,  its  length  may  be  founc  oy  (214) 


Ol  CURVED  SURFACES.  I  J5 


CHAPTER  V. 

OP    CURVED    SURFACES. 


(230)  Prop.  I.     Theorem. 
The,  equation  of  the  surface  of  a  sphere  is 

in  which  r  represents  the  radius  of  the  sphere,  m,  n,  and  p  the 
co-ordinates  of  the  centre,  and  x,  y,  and  z  those  of  any  point  in 
the  surface. 

Since  every  point  in  the  surface  of  the  sphere  is  equally  distant 
from  the  centre,  the  formula  in  (214a)  for  the  distance  between  two 
points  will  apply  to  this  case,  one  of  the  points  being  the  centre  of 
the  sphere,  and  the  other  any  point  on  the  surface,  and  the  distance 
oetween  them  the  radius  of  the  sphere.  Therefore,  by  substituting 
r  in  the  place  of  D  in  (214a),  and  fn,  n,  and  p  in  place  of  x',  y', 
and  z',  we  obtain  the  equation  required,  viz. : 

(m—xY-\'{n—yY-]r{p—zy^=r^. 

(231)  Cor,  If  the  origin  is  at  the  centre,  m,  n,  and/  disappear, 
and  the  equation  reduces  *o  3i^+y^-\-z^=^j^. 

(232)  Prop.  II.     Problem. 

To  find  the  equation  of  a  plane  tangent  to  a  sphere. 

If  a  plane  touch  a  sphere,  a  straight  line  drawn  from  the  centre 
to  the  point  of  contact  is  perpendicular  to  the  plane.  Consequently 
if  from  any  assumed  point  in  the  plane  two  lines  be  drawn,  ont  o 
the  centre  of  the  sphere  and  the  other  to  the  point  of  tane;ency. 

19 


;  t6  ANALYTICAL  GEOMETJIY 

these  two  lines,  together  with  the  radius  of  the  sphere  drawn  to  the 
point  of  tangency,  will  form  a  right-angled  triangle. 

Let  the  co-ordinates  of  the  centre  of  the  sphere  be  m,  n,  and  p  ; 
those  of  the  point  of  tangency  x',  y',  and  z' ;  and  those  of  the  as- 
sumed point  in  the  plane  x,  y,  and  z.  ■ 

By  (214tf)  we  may  obtain  the  length  of  each  side  of  the  triangle 
in  terms  of  these  co-ordinates,  viz. : 


^{m—xf-\r  (n— y)2+(7?  — %)2=the  distance  from  the  assumed  point 
to  the  centre  of  the  sphere. 

>/{m^x'Y-\-  {n—y'Y-\-  (p  — z')2=:  the  radius  of  the  sphere. 

'^{x'—xY-\r{y'—yy  +  {z'—zY—ihe    distance    from    the    assumed 
point  to  the  point  of  tangency. 

Putting  the  square  of  the  hypotenuse  equal  to  the  sum  of  the 
squares  of  the  other  two  sides,  we  obtain 

(232a)     {m-xf^-{n-yf^-(p-zf={m—x'f-^{n-yJ-\- 
{p-z'f^-{x'-xY^[y'-yf-^{z'-zf) 

which  is  the  equation  required. 

(233)  Schol.     By  expanding  the  terms  in  (232a)  and  transposing,, 
it  reduces  to 

(233a)     mx'—mx  +x'x—x''''+  ny'  —  ny+yy—y'^+pz'—pz  + 

z'z-z'^=0; 
which  may  be  expressed  in  the  form 

(2336)     {m-x)  {x'-x)-\'{n-y')  (y'_v)-f(p-z')  (z'— z)=0. 
Or, 

(233c)     (x'^m)x+{y'-n)y-\-{z'^p)z-{-{m—x')x'-\-{n-y')y'-\' 
{p-z')z'=0. 

If  we  add  (2336)  to  the  equation  of  the  sphere,  which  may  be 
expressed  in  the  form 

{m—x')  {m—x')  -\-(n—y')  (n—y')-^(p  —  z')  (p—z'y-=r^, 

we  have  the  equation  of  the  tangent  plane  expressed  in  the  fonn 
(233^)     {m-x')  {m—a)'\-(n—y')  (n-y)-f  (j9— z')  (j5-.z)  =  r* 


OF  CURVED  SUKrACli&.  }4f 

(233e)  Ex.  Determine  the  length  of  a  perpendicular  let  fall 
from  a  point  in  space,  whose  co-ordinates  x",  y",  and  z"  are  5,  7. 
and  8  miles,  upon  a  plane  which  is  tangent  to  a  sphere ;  the  co- 
ordinates of  the  centre  of  the  sphere  being  m=4  miles,  w— 2,  and 
p=5;  and  of  the  point  of  tangency,  a?'=3  miles,  y'=l,  and  z'=6.* 

Ans.     Vl2  miles. 


(234)  Prop.  III.     Theorem. 

The  equation  of  the  surface  of  a  right  cone  having  a  circular  base  is 

(m-xf-\'{n-yf-a^p-zf=0; 

in  which  m,  w,  and  p  represent  the  co-ordinates  of  the  vertex  of 
the  cone ;  x,  y,  and  z  those  of  any  point  in  the  surface ;  and  a 
the  tangent  of  the  angle  that  the  generatrix  (207)  makes  with  the 
axis  of  the  cone. 

For  convenience  we  will  suppose  the  axis  of  the  cone  to  be 
placed  parallel  to  the  axis  of  abscissas,  so  that  any  section  parallel 
to  the  horizontal  co-ordinate  plane  will  be  circular. 

Through  any  point  in  the  surface  of  the  cone  let  a  plane  be 
made  to  pass  parallel  to  the  base.  A  circular  section  will  thus  be 
formed,  the  distance  of  whose  centre  from  the  vertex  of  the  cone 
will  be  p—z;  and,  consequently,  its  radius  will  be  a{p — z).  The 
horizontal  co-ordinates  of  its  centre  will  be  the  same  as  those  of  the 
vertex  of  the  cone,  viz. :  m  and  n.  We  have,  therefore,  a  circle 
having  m  and  n  for  the  horizontal  co-ordinates  of  the  centre,  x  and 
y  for  those  of  any  point  in  the  circumference,  and  a  (p—z)  for  its 
radius.     Consequently  (172)  its  equation  is 

(m-xf^-(n-yf-a^  {p-zf=0 ; 

and  as  this  equation  is  true  of  every  circle  lying  in  the  surface  of 
the  cone,  it  is  the  equation  of  the  surface  itself 

'  The  equation  of  the  tangent  plane  in  (233c)  is  the  preferable  form  for  the 
Boluiion  of  tliia  question,  used  in  connection  with  (Sis')  and  (229). 


148  ANALYTICAL  GEOMETRY. 

(235)  Cor.  If  the  axis  of  a  cone  coincides  with  the  vertical  co- 
ordinate axis,  m  and  n  will  disappear,  and  the  equation  will  reduce  to 

x^-\-y'^—a^(p-zf=Q] 

and  if  the  vertex  be  at  the  origin,  it  will  reduce  still  farther  to 

(236)  Prop.  IV.     Problem. 
To  find  the  equation  of  the  surface  of  an  ellipsoid. 

As  the  generating  ellipse  (208)  during  its  revolution  constant!) 
lies  in  the  surface  of  the  ellipsoid,  it  is  evident  that  an  equation  tha 
represents  the  former  in  every  position,  must  represent  the  latte/ 
also. 

(236a)  Let  a^y''^-\-h^x''^  —  a%^=(i  be  the  equation  of  the  genera 
ting  ellipse ;  m,  n,  and  p  the  co-ordinates  of  the  centre ;  and  x,  y 
and  7.  those  of  any  point  in  the  surface  of  the  ellipsoid. 

In  this  and  the  two  succeeding  propositions  we  will,  for  the  sake 
of  simplicity,  suppose  the  axis  of  revolution  to  be  parallel  to  the 
vertical  co-ordinate  axis,  or  axis  of  abscissas. 

Whichever  axis  of  the  ellipse  be  taken  as  the  axis  of  revolution, 
any  point  in  the  curve  will  describe  a  circle,  the  abscissa  of  whose 
centre  will  be  z,  the  ordinates  m  and  n,  and  its  distance  from  the 
centre  of  the  ellipse  z—p.  If  it  revolve  about  the  transverse  axis, 
this  latter  distance  will  also  be  x ,  and  the  radius  of  the  circle  y' ; 
so  that  we  have  for  the  values  of  x  and  y'  the  equations 

(2366)     a:'=z-jo;     and,  by  (172),  y''^={x-mY^-{\j-nf. 

Substituting  these  values  of  x'  and  y'  into  (236a),  we  obtain  for 
the  equation 

a^  {x-mf^  a"  {y-nf^-h^  (z-py-a^b^=0. 

which  is  the  equation  required,  since  it  represents  the  ellipse  in 
every  position  during  its  revolution. 

The  equation  is  found  in  the  same  way  if  the  revolution  be  about 


OF  CURVED  SURFACES.  M9 

the  conjugate  axis,  only  that  x'  and  y'  exchange  places,  so  that  the 
equation  becomes 

(236c)        62  {x-mf-^-h^  {y-nf-\-a^  {z-py-a^b^=0, 

(237)  Cor.  1.  If  the  axis  of  revolution  coincides  with  the  ver- 
tical co-ordinate  axis,  and  the  centre  of  the  ellipse  with  the  oiigiiij 
m,  n,  and  p  will  disappear,  and  the  equations  will  become 

a2  (3^-{-y^-{-l^z'-a^b^=0 ; 


c  a-  ix 


(238)  Cor.  2.     If  b=a,  (236c)  becomes 

{x-my+  {y-ny^  (z-py-a^=^0, 

and  (237a)  becomes 

x^-{-y^+z^-a^=0; 

both  of  which  are  equations  of  the  sphere. 

(239)  SchoL  When  the  revolution  is  about  the  transverse  axis, 
the  ellipsoid  is  called  a  prolate  spheroid ;  and  when  about  the  con- 
jugate axis,  an  oblate  spheroid. 

(240)  Prop.  V.     Problem. 
To  find  the  equation  of  the  surface  of  a  paraboloid. 

As  the  generating  parabola  (208)  during  its  revolution  constantly 
lies  in  the  surface  of  the  paraboloid,  it  is  evident  that  an  equation 
that  represents  the  former  in  every  position,  must  represent  the 
latter  also. 

(240a)  Let  y'^=p'x'  be  the  equation  of  the  generating  parabola , 
m,  n,  and  p  the  co-ordinates  of  its  vertex ;  and  x,  y,  and  z  those  of 
any  point  in  the  surface  of  the  paraboloid. 

As  the  parabola  revolves  about  its  transverse  axis,  any  point  in 
tne  curve  will  describe  a  circle,  whose  radius  will  be  y',  the  abscissa 
of  its  centre  z,  the  ordinates  m  and  n,  and  its  distance  from  the 
vertex  of  the  parabola  z—p,  and  also  x'.     We  have  therefore  the 


150  ANALYTICAL  GEOMETRY. 

same  equations  as  in  (2366),  and  making  the  same  substitution  into 
(240a)  that  we  did  into  (236rt),  we  obtain  the  equation  of  the  parab- 
oloid, viz. : 

{x—my^-\-{y  —  ny^=p^  (z—p). 

(241)  Cor.  If  the  axis  of  revolution  coincides  with  the  vert'.cal 
co-ordinate  axis,  and  the  vertex  of  the  parabola  with  the  origin,  m, 
n,  and  p  will  disappear,  and  the  equation  will  become 

x^-\-lf=])'z. 

(242)  Prop.  VI.     Problem. 

To  jind  the  equation  of  the  surface  of  an  hyperholoid. 

As  the  generating  hyperbola  (208)  during  its  revolution  constantly 
lies  in  the  .surface  of  the  hyperboloid,  it  is  evident  that  an  equation 
that  represents  the  former  in  every  position,  must  represent  the 
latter  also. 

Let  a^y'^^ ^b^x''^-\- a^W  be  the  equation  of  the  generating  hyper- 
bola ;  m,  n,  and  p  the  co-ordinates  of  its  centre ;  and  x,  y,  and  z 
those  of  any  point  in  the  surface  of  the  hyperboloid. 

It  may  be  shown,  by  precisely  the  same  process  as  in  the  ellipsoid 
(236),  that  if  the  revolution  be  about  the  transverse  axis,  the  equa- 
tion will  be 

(242a)       a^  {x-mf-\-a^  {y-nf-h^  {z-pf^a%^=Q  ;     ' 

and  if  about  the  conjugate  axis, 

(2426)         a^  (z-pf-b^  {x-mf-W{y-nY-\-a%^=^Q. 

(243)  Cor.  If  we  make  the  same  supposition  as  in  (237),  equa 
tions  (242a)  and  (2426)  become 

a^  {x^^-  y^)  -  6V4-  a^b^=0 ; 
aV-62  (a;2+3/2)  ^aPb^=0. 

(244)  Schol.  1.  When  the  revolution  is  about  the  transverse 
axis,  the  hyperboloid  consists  of  two   parts   separated  from  one 


OF  CURVED  SURFACES.  151 

another,  and  it  is  called  the  hyperholoid  of  two  sheets ;  but  when 
about  the  conjugate  axis,  only  a  single  solid  is  generated,  which  is 
called  the  nyperooloid  of  one  sheet. 

(245)  Schol.  2.  The  process  employed  in  the  three  preceding 
propositions  will  srive  us  the  eauation  of  any  solid  of  revolution, 
provided  we  know  the  equation  of  tne  generating  curve 


(246)  Prop.»VIL     Problem. 

Two  given  surfaces,  plane  or  curved,  intersect :  determine  the  line 

of  intersection. 

At  the  line  of  intersection  the  co-ordinates  will  be  common  to 
both  surfaces.  We  may,  therefore,  by  means  of  the  equations  of 
the  two  surfaces,  eliminate  one  of  the  co-ordinates,  and  the  result- 
ing equation  between  two  variables  will  be  the  equation  of  the  Une 
required. 

Ex.  1.     Let  the  surfaces  be  two  planes,  whose  equations  are 
Aa;+By-|-Cz+D=0; 

Eliminating  z  between  these  equations,  we  obtain 

/A     A'\     .  /B      B'\     ./D  .  D'\  .  ^    ,. 

(c-C^J^  +  (c-d^  +  (c  +  C^)=^'    a  straight  hne. 

Ex.  2.     Let  the  equations  of  the  surfaces  be 

x^-\ry^—a\^=Q,  the  surface  of  a  cone, 
Ax+Bt/+Cz+D=0,  a  plane. 

Eliminating  z  and  uniting  terms,  we  obtain 

(B' -  ^)  y'+ 2 ABxy  +  (A2- — 2)  .7;2+ 2BDy + 2 ADx +D«=0 ; 

which  is,  by  (193),  the  equation  of  a  conic  section,  and  we  thus 
verify  the  results  which  we  obtained  geometrically  at  (57a),  (78a)t 
and  (120a). 


152  ANALYTICAL  GEOMETRY. 

(247)  Schol.  Articles  (192)  and  (193)  enable  us  to  determine,  in 
any  given  case,  whether  the  section  is  an  ellipse,  a  parabola,  an 
Hyperbola,  a  circle,  or  a  straight  line. 

Ex,  A  plane  whose  equation  is  3a;+2y-i-4«+l=0,  intersects  a 
cone  the  equation  of  whose  surface  is  a;^+y^--9z*=0;  determine 
the  line  of  intersection. 

6  24 

Ans,    An  hyperbola  whose  axes  are  --; and  ---— . 

•'^  viOl  101 


APPENDIX 


^^WW^^WVVWV\/\.'W>.>/N--V'WWVWW^\^^^VW<# 


[Seepage  18.] 


(Fig.  1— Parabola.) 


(Fig.  1— Hypobola.) 


A.     [Seepage  13.J 

1 1  is  proved  in  Bridge's  treatise  on  the  Conic  Sections,  that  il  a 
sphere  he  inscribed  in  a  cone,  so  as  to  touch  the  plane  of  any  conic 
section,  tlie  point  of  contact  is  the  focus ;  and  the  line  in  which  the 
plane  of  the  conic  •  section  intersects  that  of  the  circle,  formed  by 
the  mutual  contact  of  the  cone  and  the  sphere,  the  directrix, 

20 


154 


APPENDIX. 


Let  CDE  lepresent  a  triangular 
section  passing  through  the  vertex 
of  the  cone,  and  perpendicular 
both  to  the  plane  of  the  base  and 
the  plane  of  the  conic  section, 
cutting  the  former  in  the  nne  DE, 
the  latter  in  AB,  and  the  inscribed 
sphere  in  FGK.  Then  will  the 
point  F  be  the  focus  of  the  conic 
section,  and  the  point  H,  in  which 
the  lines  BA  and  KG  meet  when 

produced,  will  be  in  the  directrix ;  and  consequently,  AF  must  be  to 
AH  in  the  given  ratio  mentioned  in  (1). 

Produce  AB  and  ED  till  they  meet  in  T,  forming  a  triangle 
ATE  similar  to  AGH. 

Then  AG  :  AH  ::  AE  :  AT  ::  sin.  ATE  :  sin.  AET. 

But  AG=AF,  both  being  tangents  to  a  circle  from  the  same 
point. 

Therefore   AF  :  AH  ::  sin.  ATE  :  sin.  AET,   as  remarked  in 


B.     ISee  page  96.] 

Another  process,  more  strictly  analytical,  for  finding  the  equation 
of  a  tangent  to  a  circle  (and  the  same  general  method  will  apply  to 
any  other  plane  curve),  is  the  following: 

Through  P,  the  point  of 
tangency,  draw  the  line 
PP'  cutting  the  circle  in 
any  other  point  P'. 

Since  both  PT  and  PF 
pass  through  the  point  P, 
their  equations  will  be 
(156)  of  the  form 

(1) 
(2) 


y'-.y"^a'  {cc'^x") , 


APPENDIX.  165 

in  which  od  and  y'  designate  the  co-ordinates  of  the  point  P ;  x  and 
y  those  of  any  point  S  in  PT ;  x"  and  y"  those  of  any  point  in  PP', 
(and  consequently  may  represent  those  of  P')  ;  and  a  and  a  the 
tangents  of  the  angles  which  PT  and  PP'  respectively  make  with 
the  axis  of  abscissas. 

Further,  since  the  points  P  and  P'  are  both  in  the  circumference 
of  the  circle,  we  have  (173)  the  equations 

(3)  x"^  -\-y''^  -r2=0, 

(4)  x"^+ 1/^^-7^=0.   . 
Subtracting  (4)  from  (3),  we  obtain 

which  resolved  into  factors  reads 

{x'-\-x")  ix'—x")-^{y'+y")  (y'—y")=0. 

Hence     y^.y>>=J^^"n^'r^"^- 

y'-y" 

Substituting  this  value  of  y' —y"  into  (2),  and  then  dividing  bo*a 
members  by  af—x",  we  obtain 

(5)  -7+7^  =  ^- 

If  now  we  suppose  the  line  PP'  to  turn  round  the  point  P,  P' 
may  be  made  to  approach  P;  and  when  these  points  coincide,  the 
line  PP'  will  coincide  with  PT,  and  we  shall  have 
x'=oc",     y'=y'\     and  a^a'. 

Hence  (5)  will  reduce  to  the  form     - 

x' 
-=a. 

y' 

Substituting  this  value  of  a  into  (1),  we  obtain  the  equation 

x' 

y'-y=--A^'-^)' 

which  is  the  equalion  required,  and  agrees  with  that  obtained  at 


156  APPENDIX. 

C.     [Seepage  109.] 

It  affords  a  fine  illustration  of  the  beauty  of  analytical  processes 
of  investigation,  to  observe  the  changes  that  the  radius  vector  of 
an  hyperbola  undergoes  as  it  revolves  about  the  focus,  causing  w  lo 
take  different  values  from  0°  to  360°. 

By  (89a)  m^—a^=b^,  so  that  the  expressions  for  the  value  of  r  m 
(191c)  may  be  read 

b^  ,  h^ 

and  r=  — 


a— m  cos.  w  a-^m  cos.  w 

ff  we  make  w=0°,  we  have  cos.  w=l,  and  the  expressions  for  t 
in  (191c)  become 

=i—a—m,     and  r—a—m. 


a—m 

which  are  obviously  the  values  of  FA  and  FB  (Fig.  27).  Both 
being  negative  (since  m'^a),  shows  that  the  proper  radius  vector 
does  not  meet  the  curve,  but  that  produced  backward  it  meets  it  in 
tw^o  points,  viz. :  at  A  and  B. 

It  is  evident,  moreover,  that  as  the  radius  vector  revolves  both 
values  of  r  continue  negative,  vvrhile  m  cos.  w  >a ;  and  therefore 
that  up  to  that  limit  the  radius  vector  produced  backward  meets 
both  branches  of  the  curve. 

When  m  cos.  w=a,  the  two  values  of  r  become  • 

=  50,  positive ;  and  r= 


0 

the  former  of  which  may  easily  be  shown  to  be  parallel  to  the 
asymptote,  so  that  it  does  not  meet  the  curve,  and  the  latter  meets 
it  in  the  negative  direction. 

If  we  make  m  cos.  w  <  a,  and  cos.  w  positive,  as  it  will  be  while 
w<  90°,  the  first  value  of  r  becomes  positive  and  the  second  nega- 
tive, showing  that  the  radius  vector  meets  one  branch  of  the  curve 
'n  both  directions,  but  does  not  meet  the  other  at  all. 


APPENDIX  157 

If  ca=90°.  we  have  cos.  w=0,  and  the  expressions  become 

r=± =  (89«)  and  (87)  ±.\'p, 

as  was  snown  in  (1836). 

The  value  of  w  still  increasing  we  shall  have  cos.  cj  negative, 
which  will  render  the  first  value  of  r  positive,  and  the  second  nega- 
tive, so  long  as  m  cos.  w  <  a,  but  infinite  when  m  cos  w=a.  In  the 
latter  case  the  radius  vector  becomes  parallel  to  the  other  asymp- 
tote. 

If  m  COS.  w  >  a,  both  values  of  r  become  positive,  which  shows 
that  the  proper  radius  vector  meets  both  branches  of  the  curve,  but 
that  produced  backward  it  does  not  meet  either. 

When  oj=180°,  we  have  cos.  w=  — 1.  and  the  expressions  for  r 
t)ecome 


=  — a  +  m,     and    r— =a  +  w. 


a-^m  a—m 

the  same  values  as  when  w=0°,  but  with  the  opposite  sign. 

If  we  follow  the  radius  vector  through  the  two  remainm^  quad 
rants,  we  shall  find  that  the  changes  in  the  third  quadrant  correspond 
to  those  of  the  first,  and  those  in  the  fourth  to  the  second,  but  with 
the  opposite  signs. 

D.     Problem. 
To  double  a  given  cuhe.*^ 

Put  a=the  side  of  the  given  cube,  and  y=the  side  of  the  required 
cube. 

Draw  two  parabolas  having  a  common  vertex,  and  their  trans- 
verse  axes  at  right  angles  to  one  another,  the  parameter  of  one  being 
equal  to  a,  and  that  of  the  other  to  2a.  From  the  point  where  the 
D^rabolas  intersect  draw  an  ordinate  to  the  axis  of  that  which  has 

•  This  was  a  problem  of  gfreat  celebrity  among  the  ancients. 


158  .  APPENDIX 

the  greatest  parameter,  and  the  abscissa  will  be  the  side  ot  tne  re 
quired  cube. 

For  at  the  point  of  intersection  the  co-ordinates  are  common  to 
both  curves,  but  the  abscissa  of  one  is  the  ordinate  of  the  other 
and  vice  versa.     We  have,  therefore,  the  two  equations 

x'^=2ay. 
Eliminating  x  between  these  equations  and  reducing,  we  obtam 

y^=2a\ 

Cor,     If  we  eliminate  y  instead  of  x,  we  shall  obtain 

x^=4a^ ; 
so  that  the  same  construction  enables  us  to  quadruple  a  given  cube, 

E.     {Seepage  11.) 

Two  hyperbolas  so  drawn  that  the  transverse  axis  of  one  is  the 
coniugate  axis  of  the  other,  and  vice  versa,  are  called  conjugate  hy- 
terbolas ;  and  such  hyperbolas  only  have  conjugate  diameters. 


PRACTICAL  EXERCISES  IN  CONIC  SECTIONS. 

The  following  numerical  exercises  are  designed  to  render  the 
student  familiar  with  the  definitions  and  earlier  propositions  in 
Conic  Sections : 

Exercise  1.     Construct  Conic  Sections  with  the  following  ratios: 

2:3,  3:3,  5:3. 

Why  does  the  last  curve  consist  of  two  branches,  one  on  each  side 
of  the  directrix,  while  the  others  have  but  one  branch  ?  Compute 
the  transverse  and  conjugate  axes,  the  focal  distances,  and  the  prin- 
cipal parameters  of  these  curves. 

Ex.  2.  Given  (Fig.  2)  AF  =  3,  GF  =  4,  VB  =  6,  and  JRV=:5, 
to  find  AB. 

Ex.  3.  Given  (Fig.  3)  AF  =  10,  FB  =  6,  AG  =  11,  and 
HH'=2,  to  find  SS'. 

Ex.  4.  Given  (Fig.  27)  AV  =  10,  AB  =  8,  VB  =  2,  and 
K'V  =  20,  to  find  KV. 

Ex.  5.     Given  (Fig.  19)  AF  =  2,  and  FV  =  3,  to  find  EV.    . 

Ex.  6  and  7.  Given  (Figs.  2  and  27)  BV  =  2,  AV  =  20,  to 
find  the  conjugate  axis  and  principal  parameter. 

Ex.  8.  Given  the  focal  distances  of  an  ellipse  or  hyperbola  =  5 
and  20,  to  find  the  focal  ordinate. 

Ex.  9.  Given  (Fig.  35)  the  conjugate  axis  DE  =  12,  the  focal 
distance  AV  =  2,  and  VM  =  10,  to  find  FM. 

Ex.  10.     Given  (Fig.  21)  MFN  =  70°,  to  find  UMV. 

Ex.  11.  Given  (Fig.  6  or  13)  AF  =  2,  CD  =  6,  FM  =  12,  to 
find  HC  and  the  angle  FMP. 

Ex.  12.  Given  (Fig.  la,  page  12)  MN  =  3,  NP  =  9,  MF  =  5, 
NF  =  7,  PF  =  14^  to  find  KF,  AF,  and  the  transverse  axis. 


NUMERICAL   EXEEOISES 
nr 

ANALYTICAL    GEOMETRY. 


Note  to  Prop.  I,  page  85.  The  distance  OR  or  AT,  measured 
on  either  axis  from  the  origin  to  the  point  where  the  hne  crosses 
that  axis,  is  termed  an  intercept ;  when  neither  axis  is  specified,  the 
intercept  is  understood  to  be  taken  on  the  axis  of  ordinates. 

The  angle  which  a  line  makes  with  the  axis  of  abscissas  is  called 
the  slope  of  the  line ;  thus,  in  Fig.  58,  XTP  is  the  slope  of  the  line 
PT.  It  is  always  measured  from  the  ri^ht  hand  to  the  point  of 
intersection  of  the  given  line  with  the  axis  of  abscissas,  and  thence 
upwards  :  consequently,  of  the  four  angles  made  by  the  line  meet- 
ing the  axis  of  abscissas,  the  upper  right-hand  angle  is  meant.  The 
slope  may  therefore  be  any  angle  less  than  180°.  If  the  line  is 
parallel  to  the  axis  of  abscissas,  the  slope  is  0°. 

The  position  of  a  point  may  be'  briefly  designated  by  placing  in 
a  parenthesis  the  value  of  its  abscissa  followed  by  that  of  its  ordi- 
nate; thus,  the  point  (3,  —5)  is  a  point  whose  abscissa  is  +3,  and 
its  ordinate  —5.  If  the  point  were  on  one  of  the  axes,  thus  making 
one  of  its  co-ordinates  disappear,  zero  will  take  its  place ;  thus,  the 
point  (0,  —4)  refers  to  a  point  on  the  axis  of  ordinates  at  a  distance 
of  4  below  the  origin. 

Exercise  1.  Prove  Proposition  I  when  the  point  P  is  in  the 
second  angle,  and  the  line  passes  through  the  second,  first,  and 
fourth  angles.  [In  this  case,  AD  =  —x,  the  tangent  of  PTD 
remains  equal  to  -\-a  (as  it  always  is),  but  this  angle  being  now 
obtuse  by  the  conditions  of  the  construction,  its  supplement  will  be 


NUMERICAL    EXERCISE3.  161 

employed  in  the  right-angled  triangle,  and  its  tangent  will  equal 
—a',  according  to  the  principle  in  Trigonometry  that  the  tangent 
of  any  angle  and  that  of  its  supplement  have  contrary  signs.]  The 
resulting  equation  will  be  of  the  same  form  as  given  in  the  Proposi- 
tion, viz.,  y:=ax-\-'b,  showing  that  the  introduction  of  negative 
quantities  has  not  affected  the  result. 

Ex.  2.  Write  the  equation  of  a  straight  line,  when  the  values 
of  a  and  b  are  given  ;  for  example,' a  =  — 2,  J  =  +3.  What  is  the 
slope  9    Is  it  an  obtuse  or  an  acute  angle  ? 

Ex.  3.  Write  the  equation  of  a  straight  line  when  oj  =  —  J, 
and  J  =  +2^.     What  is  the  slope  ?    What  are  the  intercepts  ? 

Ex.  4.  In  what  angle  is  the  point  (—9,  +4)  ?  In  what  angle 
is  ihQ  point  (5,  — 17)  ?    In  what  angle  is  ihQ  point  (—3,  — 5)  ? 

Ex.  5.  Prove  by  drawing  that  a  straight  line  will  pass  through 
the  three  points  named  in  the  last  exercise. 

Ex.  6.  Prop.  II,  page  86.  Write  the  equation  of  the  line  pass- 
ing through  the  point  (4,  —7),  and  making  an  angle  of  20°  with 
the  axis  of  abscissas. 

Ex.  7.  Through  what  given  poi?it  does  a  line  pass  whose  equa- 
tion is  2  —  y  =  0.7002  (— 5  —  a;).  In  which  angle  is  the  given 
point?  What  is  the  slope  of  the  line?  How  would  the  line  be 
drawn,  if  its  equation  were 

%—y=  -0.7002  (-5 -a;)? 

Ex.  8.  Write  the  equation  of  a  line  passing  through  the  point 
(  +  7,  —2)  and  making  an  angle  of  50°  with  the  axis  of  abscissas. 

Ex.  9.  Prop.  Ill,  page  87.  Write  the  equation  of  the  line 
that  passes  through  the  point  B  =  (—4,  —3)  and  C  =  (-|-4,  —2). 
Through  what  angles  does  it  pass  ?  Determine  approximately,  by 
drafting,  the  slope  of  the  line. 

Ex.  10.  Given  the  abscissa  of  a  third  point,  D  =  24,  in  the 
line  in  the  last  exercise,  what  is  the  corresponding  ordinate  ? 

Ans,  y  =  i. 


162  NUMEEICAL    EXERCISES. 

Ex.  11.  Form  the  equation  of  a  line  passing  through  the  points 
(~5,  +2)  and  (6,  —3),  and  reduce  it  to  the  form 

lly  =  —5a;  — 3. 

Ex.  12.  Draft  the  triangle  whose  sides  are  represented  by  the 
equations: 

1.     y  =  ix-\-3, 

3.     y  =  ix-l. 

Ex.  13.  What  figure  is  that  whose  angles  are  A  =  (2,  3), 
B  =  (2,  8),  0  i=  (7,  8),  and  D  =  (7,  3).  Draft  it.  What  are  its 
angles  ? 

Ex.  14.     What  figure  is  enclosed  by  sides  whose  equations  are 
1.  2y  +  2x  =z3x-i-3  +y, 

3.  sy^2x^6  =  ^^^  +  l, 

4.  cc-\-y  =  —  3.* 

Ex.  15.  The  vertices  of  a  triangle  are  (2,  8),  (--6, 1),  and 
(0,  —4).    Required  the  equations  of  the  sides. 

Ex.  16.  Prop.  V,  page  88.  Find  the  distance  between  the 
points  (—7,  +2)  and  (  +  17,  —5).  Ans.  ±  25. 

Ex.  17.  Find  the  distance  between  the  points  (—7,  —1)  and 
(  +  5,  +4).  Ans.  ±13. 

Ex.  18.  Find  the  distance  from  the  origin  to  the  point 
(__6,  -8).  Ans.  ±  10. 

*  From  Olney's  General  Geometry. 


NtTMEBICAL    EXERCISES.  163 

Ex.  19.     Prop.  VI,  page  89.    Determine  the  angles  of  the  tri- 
angle ABC  formed  by  the  lines  whose  equations  are 

AB,  y  =  ix- 13, 
BC,         ^/==ix-^, 

AC,  l/  =  2x-^, 

Ans.  A  =  12°  14',  B  =  45°,  0  =  122°  45'. 

Ex.  20.     At  what  angle  do  these  lines  meet  ? 

02;  —  4^  —  52  =  0,  • 

•       2/  +  4  =  i(^-3). 

Ans.  46°. 

Ex.  21.     At  what  point  do  the  lines  named  in  the  last  example 
meet  ?    [See  Prop.  XVIII,  page  119.]       Ans.  (  +  7.61,  -3.49). 

Ex.  22.     At  what  point  do  these  lines  intersect  ? 

3a;  4.  2//  —  12  =  0, 

4a;  +  3y  —  17  =  0. 

Ans,  (2,  3). 
Ex.  23.     At  what  point  do  these  lines  intersect  ? 

|  +  22/-5  =  0, 
2a;  — 1 

y ^  =  1. 

Ans.  (3,  2). 

Ex.  24.    At  what  point  do  these  lines  intersect  ?    [See  p.  95.] 

lOy  =  8rc2  +  5, 

1/  =  6x  —  8. 
Ans.  They  approach  each  other  closely,  bnt  do  not  meet. 

Ex.  25.     At  what  points  do  the  curves  on  page  95,  Examples  4 
and  7,  intersect  ?    Also  Examples  3  and  6  ? 


164  NUMERICAL    EXERCISES. 

Ex.  26.     At  what  angle  do  these  lines  intersect  ? 
—  5a;  +  4«/  +  14  =  0, 

Sx  —  ei/  =  '7ix  +  1. 


Ans.  45°. 


Ex.  27.     What  angles  do  the  following  lines  make  with  each 
other? 

BO,        y  =  li^  -  1  J, 

AC,        y  =  -^x-^  30, 
AB,        ^  =  —  Hx  —  14. 

Ex.  28.     Where  are  the  vertices,  and  what  are  the  angles  of  the 
triangle  represented  by  the  equations : 

1.  6y  =  7x  —  9, 

2.  9y  =  —  4:X  +  30, 

3.  3y  =  —  llic  —  42. 

Ex.  29.     Prove  that  these  lines  meet  at  an  angle  of  45° : 
6x  —  4:ij  =  52, 

y  +  4  =  i(T-3). 

Ex.  30.     Prove  that  these  lines  meet  at  an  angle  of  45°.! 
y  =  Sx  —  1l, 

y  =  —2x  -\-  5. 

^      Ex.  31.     Prove  that  these  lines  are  parallel: 

2y  +  2x  =  Sx  +  3  -\-  y. 

—L-l —  2x  ^  6  —  y. 

2  ^ 

Ex.  32.     Form  the  equation  of  a  line  passing  through  the  origin 
and  perpendicular  to  the  line  whose  equation  is 

y  =  -ix-5. 

Ex.  33.     Find  the  distance  from  the  origin  to  the  line 

y=-ix^5. 

Ans.   ±  4. 


KUMERICAL    EXERCISES^  165 

Ex.  34.     Find  the  distance  from  the  point  (6,  8)  to  the  line 
y  =  —Ix  —  b. 

Ex.  35.     Find  the  distance  from  the  origin  to  the  line 

y  =  —  2a;  +  5. 

Ans.   ±  ^/b. 

Ex.  36.  In  the  triangle  ABC  (Ex.  27),  what  is  the  length  of 
AB  ?  What  is  the  distance  from  C  to  AB  ?  What  is  the  area  of 
the  triangle  ? 

Ex.  37.  Draw  the  triangle  MPR,  whose  sides  are  represented 
by  the  following  equations,  and  compute  its  area : 

MP,  y  =  ^x-4., 

MR,        y  =  -i.r  +  5, 
PR,         y  =  :^x-%. 

Ans.  Representing  by  H  the  foot  of  the  perpendicular  drawn 
from  M  to  the  side  PR,  we  have 

M  =  (4«,3A),  P  =  (^l|,  _6|), 

R  =  (184,  -  ^\  H  =  (6H,  -  4^), 

PR  =  21.08,  MH  =  7.89, 

Area  =  83.22. 

Ex.  38.  Find  the  area  of  the  triangle  ABC,  when  the  co-ordi- 
nates of  A  are  (2,  3),  of  B  (4, 1),  and  of  C  (5,  6). 

Ans.  Area  =  6. 

Ex.  39.  Find  the  area  of  a  triangle  DEP,  when  the  co-ordinates 
are 

D  ^  (5,  10),        E  =  (17, 10),        F  =  (11,  4). 

Ex.  40.  Find  the  area  of  the  quadrilateral  BCDE,  having 
given 

B  =  (-4,  7),  C  =  (6,  3), 

D  =  (2,  -1),  E  =  (-3,  -2). 

Ans.  51. 


166 


NUMERICAL    EXERCISES. 


Ex.  41.  Chapter  II,  Prop.  II.  Prove  the  equation 
(x  —  mf  +  (2/  -  nf  —  r2  =  0, 
when  the  figure  is  so  constructed  that  both  the  centre  0  and  the 
point  P  are  in  the  third  angle,  thus  causing  the  quantities  that  enter 
into  the  equation,  excepting  r,  to  become  negative.  [Note,  r  not 
being  necessarily  a  horizontal  nor  a  vertical  line,  may  always  be 
regarded  as  positive.] 

Ex.  42.  Prove  the  same  equation  when  the  circumference  lies 
partly  in  each  of  the  four  angles,  and  the  points  C  and  P  are  in 
separate  angles. 


Prop.  III.    Theorem. 
To  be  added  to  page  96. 
The  equation  of  a  tangent  to  a  circle  is 

in  which  m  and  n  denote  the  co-ordinates  of  the  centre,  x^  and  y' 
those  of  the  point  of  tangency,  and  x  and  y  those  of  any  other  point 
in  the  tangent. 

T 


Let  0  be  the  centre  of  the  circle,  M  any  point  on  the  circumfer- 
ence, MT  the  tangent  to  the  circle,  and  P  any  point  in  the  tangent. 

Let  AH==m,  CH  =  w,  AN  =  a;',  MN  =  y',  AD  =  a;,  and 
PD  =  y.     Since  the  line  MP  passes  through  the  given  point  M  and 


NTJMEEIOAL    EXEKCISES.  167 

the  required  point  P,  its  equation  will  be  (Prop.  IE,  page  86)  of  the 
form, 

y'  —y  =  a{x'  ^x), 

m  which  a  represents  the  tangent  of  the  angle  MTN.  Draw  CE 
perpendicular  to  MK  The  triangle  MEC  is  similar  to  the  triangle 
MTN,  and  hence  the  angle  OME  equals  the  angle  MTN.  By  trig- 
onometry, the  tangent  of  the  angle  CME  =  ^^  =  -7-— —  =  a. 

Hence,  y' -  ^^  =  ^^  (a;' -  a;). 

Note.  Prop.  ITT,  Problem,  may  be  considered  a  corollary  to  this 
theorem. 

QUESTION'S.  Where  is  the  point  of  tangency  when  m  —  a/  =  0  ? 
Where  is  it  when  ^'  —  w  =  0  ?  What  is  the  slope  of  the  tangent  in 
these  two  cases  ?  < 


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